Introduction to Mathematical Logic - IMJ-PRGadrien.deloro/teaching... · ·...
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Introduction to Mathematical Logic Adrien Deloro Spring 2009
Transcript of Introduction to Mathematical Logic - IMJ-PRGadrien.deloro/teaching... · ·...
Introduction to Mathematical Logic
Adrien Deloro
Spring 2009
Contents
᾿Εν ἀρχῇ ἦν ὁ λόγος 4
0 Propositional Logic (Allegro) 70.1 Syntax . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
0.1.1 The Language of Propositional Logic . . . . . . . . . . . . 80.1.2 Well-Formed Formulas . . . . . . . . . . . . . . . . . . . . 80.1.3 Unique Readability . . . . . . . . . . . . . . . . . . . . . . 9
0.2 Semantics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110.2.1 Truth Assignments . . . . . . . . . . . . . . . . . . . . . . 110.2.2 Satisfaction and Semantic Consequence . . . . . . . . . . 120.2.3 Simplifying the Language . . . . . . . . . . . . . . . . . . 13
0.3 Natural Deduction (in Classical Logic) . . . . . . . . . . . . . . . 150.3.1 Syntactic Consequences and Deductions . . . . . . . . . . 160.3.2 Our First Deductions . . . . . . . . . . . . . . . . . . . . 180.3.3 Simplifying the Language (and Presentation) . . . . . . . 200.3.4 The Soundness Theorem . . . . . . . . . . . . . . . . . . . 27
0.4 The Completeness Theorem . . . . . . . . . . . . . . . . . . . . . 280.4.1 Extending the Theory . . . . . . . . . . . . . . . . . . . . 290.4.2 Finding a Truth Assignment . . . . . . . . . . . . . . . . 30
0.5 The Compactness Theorem . . . . . . . . . . . . . . . . . . . . . 310.5.1 Compactness . . . . . . . . . . . . . . . . . . . . . . . . . 310.5.2 Applications to Decidability* . . . . . . . . . . . . . . . . 320.5.3 A Topological Proof* . . . . . . . . . . . . . . . . . . . . . 34
0.6 A Little Modal Logic* . . . . . . . . . . . . . . . . . . . . . . . . 350.6.1 Semantics* . . . . . . . . . . . . . . . . . . . . . . . . . . 350.6.2 Proof Theory* . . . . . . . . . . . . . . . . . . . . . . . . 360.6.3 Completeness* . . . . . . . . . . . . . . . . . . . . . . . . 38
0.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
1 First-Order Logic (Largo) 451.1 Syntax . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
1.1.1 First-order Languages . . . . . . . . . . . . . . . . . . . . 461.1.2 Terms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 471.1.3 Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
1.2 Semantics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
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1.2.1 Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . 501.2.2 Parameters and Interpretations . . . . . . . . . . . . . . . 511.2.3 Satisfaction and Semantic Consequence . . . . . . . . . . 51
1.3 Substitutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 521.3.1 Substitutability . . . . . . . . . . . . . . . . . . . . . . . . 531.3.2 A Renaming Algorithm . . . . . . . . . . . . . . . . . . . 541.3.3 Substitutions and Satisfaction . . . . . . . . . . . . . . . . 55
1.4 Deductions and Soundness . . . . . . . . . . . . . . . . . . . . . 571.4.1 Deductions . . . . . . . . . . . . . . . . . . . . . . . . . . 571.4.2 Simplifying the Language . . . . . . . . . . . . . . . . . . 581.4.3 Soundness . . . . . . . . . . . . . . . . . . . . . . . . . . . 61
1.5 Completeness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 621.5.0 Strategy and Witnesses . . . . . . . . . . . . . . . . . . . 631.5.1 Expanding the Language . . . . . . . . . . . . . . . . . . 641.5.2 Extending the Theory . . . . . . . . . . . . . . . . . . . . 661.5.3 Finding a Structure (and Assignment) . . . . . . . . . . . 68
1.6 Consequences and Compactness . . . . . . . . . . . . . . . . . . 701.6.1 Decidability* . . . . . . . . . . . . . . . . . . . . . . . . . 701.6.2 Compactness . . . . . . . . . . . . . . . . . . . . . . . . . 711.6.3 Non-Standard Analysis* . . . . . . . . . . . . . . . . . . . 72
1.7 An Alternate Proof of Compactness* . . . . . . . . . . . . . . . 741.7.1 Filters and Ultrafilters* . . . . . . . . . . . . . . . . . . . 741.7.2 Ultraproducts (The Łoś Structure)* . . . . . . . . . . . . 761.7.3 Alternate Proof of Compactness* . . . . . . . . . . . . . . 78
1.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78
2 Second-Order Logic (Presto) 852.1 Compactness fails . . . . . . . . . . . . . . . . . . . . . . . . . . 862.2 Peano Arithmetic . . . . . . . . . . . . . . . . . . . . . . . . . . 87
1’ Some Model Theory (Furioso) 891’.0 A Word on Model-Terrorists . . . . . . . . . . . . . . . . . . . . 891’.1 Elementary Equivalence; Inclusion and Elementary Inclusion . . 92
1’.1.1 Elementary Equivalence . . . . . . . . . . . . . . . . . . . 921’.1.2 Inclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . 931’.1.3 Elementary Inclusion . . . . . . . . . . . . . . . . . . . . . 95
1’.2 Morphisms, Elementary Morphisms, Categoricity . . . . . . . . . 961’.2.1 Morphisms . . . . . . . . . . . . . . . . . . . . . . . . . . 971’.2.2 Elementary Morphisms . . . . . . . . . . . . . . . . . . . 981’.2.3 Categoricity . . . . . . . . . . . . . . . . . . . . . . . . . . 99
1’.3 Löwenheim-Skolem Theorems . . . . . . . . . . . . . . . . . . . . 1001’.3.1 The Substructure Generated by a Set . . . . . . . . . . . 1001’.3.2 Skolem Functions and the Descending Version . . . . . . . 1011’.3.3 The General Version and the Łoś-Vaught Criterion . . . . 102
1’.4 Back-and-Forth Methods . . . . . . . . . . . . . . . . . . . . . . 1031’.4.1 Dense Linear Orderings . . . . . . . . . . . . . . . . . . . 104
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1’.4.2 ∞-isomorphisms . . . . . . . . . . . . . . . . . . . . . . . 1061’.4.3 Finitary Back-and-Forth* . . . . . . . . . . . . . . . . . . 108
1’.5 Quantifier Elimination and ω-Saturation . . . . . . . . . . . . . . 1111’.5.1 Elimination Sets . . . . . . . . . . . . . . . . . . . . . . . 1111’.5.2 ω-Saturated Models . . . . . . . . . . . . . . . . . . . . . 1141’.5.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . 118
1’.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123
Indices 128Index of Notions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128Index of Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130
List of Lectures 131
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᾿Εν ἀρχῇ ἦν ὁ λόγος
The Greeks didn’t have a formal language, which in Europe appears duringthe Renaissance. Because of our education it now seems very natural to usthat every mathematical statement may be translated into specific signs; yetthis fascinating fact has deeply moved Leibniz’ mind. The adequation of sucha poor alphabet to the manifold hues of mathematics is striking indeed. Oneshould however not succumb, and their temptations are always present, to theSirens of sheer formalism; a mathematician aims at transcribing his intuitionsinto symbols, not at assembling symbols and then seeing if they by chance makeany sense. Never does he renounce lucid thinking nor his mother language; andeven less does the logician.
A logician is a mathematician who bears in mind both the mathematicaltruths and the way they are stated; able to distinguish the meaning from theformulation, he can study their interplay. If the mathematician regards languageas a tool to express some interesting properties, the logician thinks the propertiesof this tool are themselves worth the interest.
Logic is the study of the relations between things and wordsIn any particular form of logic, we need both a system of objects and a language.We must of course make the language expressive enough for the properties ofthe objects which appear relevant to us, but not more.
An example will shed some light.• We are all familiar with the encoding of numbers in the standard, decimal,number system. Our objects here are numbers; the language is the collec-tion of representations of numbers. The latter is better described by itsalphabet (the symbols 0 through 9), and the formation rules of what is anacceptable representation of a number. There is only such rule: writingfrom the right to the left, the last digit is not 0.
• Once this is done, we need to be able to translate back and forth. Givena number, it is easy to compute recursively (using remainders modulo 10etc.) its decimal representation. Conversely, the number can be retrievedfrom its expansion via a certain sum. We have thus seen that our lan-guage of representations of numbers in base 10 is sufficient to describe allnumbers.
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Though the objects and language might seem a little less familiar, we shallfollow essentially the same guidelines in our study of various forms of logic:
• agree on a collection of interesting objects
• give a set of basic symbols (an alphabet)
• give a collection of rules for assembling them (a syntax)
• give a meaning to the language (a semantics).
After that, the study of the relationships between the language and what itstands for can start.
Logic turns proofs into objectsThe expressive power of a language may be measured by the notions of conse-quence it creates; let us explain why consequence is essentially twofold.
The main feature of contemporay mathematical logic is that not only mathe-matical statements may be made formal, but also those particular combinationsof mathematical statements which are called proofs. In a sense, modern logic isable to treat mathematical sentences themselves as words of a language whosesentences are mathematical deductions. In this case the grammar consists ofthe rules explaining which conclusions may be drawn from given assumptions.(Treating proofs as formal objects is essentially as shocking as treating intuitiveproperties as formal statements; in a sense, the XIXth and XXth centuries werethe Renaissance of Mathematical Logic.)
On the semantic side, the concept of logic consequence is that of entail-ment. Similar to material implication which is only about the truth value of itscomponents, this semantic approach of consequence is concerned only with oneproperty never occuring without another one.
Hence there are two notions of consequence:
• consequence at a syntactic level: ϕ follows from Σ if there is a proof, usingΣ, of ϕ. This is written Σ ` ϕ, which may be read “Σ proves ϕ”.
• consequence at a semantic level: ϕ follows from Σ if it is the case thattruth of Σ always causes truth of ϕ. This is written Σ |= ϕ, which maybe read “Σ entails ϕ”.
A syntax remarkably fits a semantics when the two notions ` and |= coincide.This adequation holds in Propositional Logic and First-Order Logic, but failsin Second-Order Logic, which is in a sense too expressive. These three forms oflogic are explained hereafter.
The semesterIn the course of the semester we shall study three levels of logic, from the lessto the more expressive.
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• Propositional Logic concerns itself with compound propositions, using thefive usual connectives. In propositional logic, our basic symbols alreadyrepresent propositions; we connect them with the usual connectives. Thisform of logic can be dealt with using truth tables: it suffices to specify foreach P is P is true or false. It is not very expressive, but this playgroundis an extremely interesting field of investigation for dgging up notionsrelevant in deeper logics.
• First-order logic is the natural language for elements. Our basic symbolsare now for elements, and certain (given) relations they can have. Onesees that this is already deeper than propositional logic; one also sees thatone has to introduce quantification on elements. First-order logic is strongenough to encode much of interesting mathematics, and still enjoys verynice properties.
• Second-order logic builds on the previous one. Our symbols can now alsobe for “predicates‘”, that is for sets, relations. But unlike first-order logic,second-order logic may quantify on those. So we need new quantifiers(roughly speaking, we now may quantify on classes, not only on objects).Second-order logic is extremely expressive, and this is actually the reasonwhy the beautiful, deep theorems we shall prove in the first two cases willfail in this somehow too general context.
Looking back at our first steps, we realize that propositional logic is somekind of zeroeth-order logic. After the investigation of these three forms of logic,we should believe that the one that best fits our needs is first-order logic. Weshall then move to model theory, which is the mathematical study of first-ordertheories. This should explain the rather unorthodox numbering.
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Chapter 0
Propositional Logic(Allegro)
Propositional logic, sometimes called Propositional Calculus, is interested incompound propositions; it is the logic which underlies truth tables. In spite ofits little mathematical interest, it is a perfect field for a first encounter withimportant notions.
We shall introduce a formalization of deduction in the setting of propositionallogic, and study (some of) its properties. This will be a brief account of prooftheory.
We shall also introduce the objects of propositional logic, that is truth as-signments. These are on the side of meanings, called semantics. They bringtheir own notion of consequence; the main result of this chapter which servesas a training area for more expressive logics, is that in propositional logic, bothnotions coincide.
In this chapter:
• Describe the language of propositional logic (§0.1)
• Describe the objects of propositional logic (§0.2)
• Formalize the notion of entailment (§0.2.2)
• Formalize the notion of deduction in propositional logic (§0.3)
• Show that entailment and deduction coincide (§0.3.4 and 0.4)
• Derive the compactness theorem and applications (§0.5)
• An incursion to modal logic* (§0.6)
Lecture 1 (Language and Wff’s; Unique Readability)
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0.1 SyntaxWe first deal with the familiar syntax of propositional logic. The expressions are(finite) sequences of well-known symbols (§0.1.1), but not all such expressionsmay be given a meaning. This will lead us to the relevant notion of formula(§0.1.2). A key, and fairly natural, property of the language is that there isexactly one way to read a relevant formula (Theorem 0.1.9 of §0.1.3).
0.1.1 The Language of Propositional LogicAs we have said, propositional logic is interested in building coumpound propo-sitions from elementary propositions.
Definition 0.1.1 (language of propositional logic). The language of proposi-tional logic consists of:
• two punctuation symbols : the parentheses “(” and “)”
• five connectives ¬, ∧, ∨, →, ↔
• a set A of sentence symbols A1, . . . , An, . . . .
One should already have an intuition that ¬ and → suffice to encode theother three connectives. This unformal idea will be given a precise meaning in Exercise 0.1§0.3.3.
Definition 0.1.2 (expression). An expression is any (finite) sequence of sym-bols from the alphabet. Its length is its length as a sequence.
Remark 0.1.3. The set E of expressions of propositional logic is countable.
Technically, Definition 0.1.1 defines the alphabet of propositional logic. It isnever perfectly clear what “language” means: does one mean the alphabet, theset of expressions, or the set of well-formed formulas (Definition 0.1.4 below)?Very fortunately, the cardinals of the alphabet, the set of expressions, and theset of well-formed formulas, are the same.
0.1.2 Well-Formed FormulasAs “(→ A1¬” is an expression, we must clarify the formation rules.
Definition 0.1.4 (well-formed formula). The collection WFF of well-formedformulas (wff in short) is the smallest set such that:
• each An is a wff;
• if ϕ and ψ are wff, so are (¬ϕ), (ϕ ∧ ψ), (ϕ ∨ ψ), (ϕ→ ψ), and (ϕ↔ ψ).
Remark 0.1.5. As this is the only relevant notion of formula anyway, we shallsoon omit “well-formed”, and say “formula”, for short.
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Definition 0.1.6 (theory). A theory is a set of wff’s.
Wff’s are built from the sentence symbols with connectives. As this is arecursive definition, most of our proofs on formulas will proceed by induction(notice that the length of a compound proposition is bigger than that of thepropositions which compose it).
Let us be more technical and check that Definition 0.1.4 makes sense.
Notation 0.1.7. We define the following functions.
C¬ : E → Eϕ 7→ (¬ϕ)
C∧ : E2 → E C∨ : E2 → E(ϕ,ψ) 7→ (ϕ ∧ ψ) (ϕ,ψ) 7→ (ϕ ∨ ψ)
C→ : E2 → E C↔ : E2 → E(ϕ,ψ) 7→ (ϕ→ ψ) (ϕ,ψ) 7→ (ϕ↔ ψ)
Notice that these functions actually take WFF (resp. WFF2) to WFF.
Notation 0.1.8. Let WFF0 = A, and for n ∈ N,
WFFn+1 = WFFn⋃C¬(WFFn)
⋃C∧(WFF2
n)⋃· · ·⋃C↔(WFF2
n)
It is clear that for all n ∈ N, WFFn is countable. It is also clear that WFF =∪n∈N WFFn. This proves that WFF is a set, and even a countable set. Moregenerally, if A is infinite (but perhaps uncountable), then Card WFF = CardA.
0.1.3 Unique ReadabilityThere is still something which needs to be proved about our well-formed formu-las: that no ambiguity arises when we read one.
Theorem 0.1.9 (unique readability). The restrictions to WFF of the functionsC¬, C∧, C∨, C→, and C↔ are injective and have disjoint images.
Corollary 0.1.10. Let ϕ be a wff of length > 1. Then there is a unique decom-position of ϕ among the following possibilities:
(¬ψ1), (ψ1 ∧ ψ2), (ψ1 ∨ ψ2), (ψ1 → ψ2), (ψ1 ↔ ψ2)
(This means that the connective and the composing propositions are unique.)
An initial segment of an expression E is an expression E1 such that thereis an expression E2 with E = E1E2 (concatenation). A proper initial segmentE1 of E is an initial segment such that E1 6= E. Notice that the empty (blank)sequence is an expression but not a wff.
The proof of Theorem 0.1.9 will rely on the following key observation.
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Lemma 0.1.11 (balanced parenthesing).
(i). A wff has as many left parentheses as right parentheses.
(ii). A non-empty proper initial segment of a wff has strictly more left paren-theses than right parentheses.
(iii). No proper initial segment of a wff is a wff.
Proof .
(i). We show the property by induction on (the length of) ϕ, using the de-scription of WFF found above (Notation 0.1.8).If ϕ ∈ WFF0, then ϕ is a sentence symbol, and this is obvious. Assumethe result is know for elements of WFFn. Then an element of C¬(WFFn)is of the form ϕ = (¬ψ) with ψ ∈WFFn, so by induction ψ has as manyleft parentheses as right parentheses, and so does ϕ. The same holds forelements of C∧(WFF2
n), C∨(WFF2n), C→(WFF2
n), and C↔(WFF2n). So any
element of WFFn+1 has as many left parentheses as right parentheses.By induction, the property is true on WFF.
(ii). Same method. If ϕ ∈ WFF0, then ϕ is a sentence symbol, so it has nonon-empty proper initial segments, and the claim is true!Assume it holds for elements of WFFn. Let ϕ be of the form (¬ψ), forψ ∈ WFFn. Consider a non-empty proper initial segment ϕ1 of ϕ: weshow that ϕ1 has more (’s than )’s.
• If ϕ1 is “(” or “(¬”, we are done.• If ϕ1 is “(¬ψ1” where ψ1 is a non-empty proper initial segment of ψ,then we are done by induction.
• If ϕ1 is “(¬ψ” then we are done as ψ has as many (’s as )’s by (i).• There are no other cases, as ϕ1 is a proper initial segment of ϕ.
This shows the property for ϕ. One copes with other four connectives ina similar vein.
(iii). A consequence of (i) and (ii) (the empty expression is not a wff anyway).Exercise 0.2
Proof of Theorem 0.1.9. Assume that C�(ϕ1, ψ1) = C�(ϕ2, ψ2), that is:
(ϕ1 � ψ1) = (ϕ2 � ψ2)
for well-formed formulas ϕ1, ψ1, ϕ2, ψ2 and connectives � and �.We may delete the first “(”, getting
ϕ1 � ψ1) = ϕ2 � ψ2)
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Then ϕ1 or ϕ2 must be an initial segment of the other; without loss ofgenerality ϕ1 is an initial segment of ϕ2. If ϕ1 is a proper initial segment ofthe wff ϕ2, then from balanced parenthesing (Lemma 0.1.11 (iii)), ϕ1 can’t bea wff, a contradiction.
So ϕ1 = ϕ2, and deleting it we reach
�ψ1) = �ψ2)
At this point it is clear that � = �; deleting it, and the final right parentheses,we derive
ψ1 = ψ2
Of course the case of C¬ is even easier than that of binary connectives, soeverything is proved.
Such a quick proof of the Unique Readability Theorem is the reason why wehave introduced so many parentheses, in spite of the disgusting redundanciesthey create. From now on we freely omit parentheses when there is no ambiguity,i.e. we write ¬ϕ ∧ ψ instead of ((¬ϕ) ∧ ψ).
End of Lecture 1.
Lecture 2 (The Semantics of Propositional Logic)
0.2 SemanticsWe briefly escape the claws of syntax, and give sense to our symbols. So farthey bore no meaning; at no point yet did we agree on what they stand for.
0.2.1 Truth AssignmentsLet us establish the connection between the language and the objects it is about.Due to the relative poverty of propositional logic, not much can be said: ev-erything is described once each sentence symbol of A will have been assigned atruth value.
Definition 0.2.1 (truth values). There are two truth values: True (T ) andFalse (F , also denoted ⊥).
Notice that everything from now on could be defined with more truth values;for instance continuous logic investigates the case where the set of truth valuesis [0, 1]. But the important theorems we want might no longer hold.
Definition 0.2.2 (truth assignment). A truth assignment is a function v : A →{F, T}.
Once the sentence symbols have been given a truth value, the “natural”truth value of every wff is known.
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Proposition 0.2.3. Let v be a truth assignment. Then there is a unique ex-tension v : WFF→ {F, T} such that:
• v(An) = v(An) for each n ∈ N;
• v(¬ψ) = T iff v(ψ) = F ;
• v(ψ1 ∧ ψ2) = T iff v(ψ1) = T and v(ψ2) = T ;
• v(ψ1 ∨ ψ2) = T iff v(ψ1) = T or v(ψ2) = T ;
• v(ψ1 → ψ2) = T iff v(ψ1) = F or v(ψ2) = T ;
• v(ψ1 ↔ ψ2) = T iff v(ψ1) = v(ψ2).
Proof . This extension v is well-defined by Unique Readability (Theorem 0.1.9).A quick induction shows that it is unique.
Example 0.2.4. If v(A1) = v(A2) = T and v(A3) = v(A4) = F , then one hasv((A1 → ¬A2)↔ (A3 ∨A4)) = T .
Only a little later (at the end of §0.2.3) shall we identify v to is extension v.
0.2.2 Satisfaction and Semantic ConsequenceDefinition 0.2.5 (satisfaction). Let Σ be a theory and v be a truth assignment.v satisfies Σ if for all ϕ ∈ Σ, v(ϕ) = T .
Example 0.2.6.
• v(A1) = F , v(A2) = T satisfies (A1 → A2).
• v(An) = T for all n ∈ N satisfies {A1, A1 ∧A2, . . . , A1 ∧ · · · ∧An, . . .}.
Definition 0.2.7 (satisfiability). Let Σ be a theory. Σ is satisfiable if there isa truth assignment satisfying it.
Example 0.2.8.
• {¬A1, A1 → A2,¬A1 → A3} is satisfiable (try for instance the truthassignment v(A1) = v(A2) = F , v(A3) = T ).
• {A1 → ¬A1} is not satisfiable.
• {A1,¬A2, A1 ↔ A2} is not satisfiable. Exercise 0.3
A very natural question occurs: when is a theory satisfiable? it will have ananswer with the Compactness Theorem, Theorem 0.5.2 below.
The most important notion of logic is the following.
Definition 0.2.9 (semantic consequence). Let Σ be a theory and ϕ a wff. Σentails ϕ, denoted Σ |= ϕ, if every truth assignment satisfying Σ satisfies ϕ.
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One also says that Σ tautologically implies ϕ. If ∅ |= ϕ, one may then saythat ϕ is a tautology, or that ϕ is valid. We do not insist on classical terminology.
Lemma 0.2.10. Σ ∪ {ϕ} is satisfiable iff Σ 6|= ¬ϕ.
Proof . Suppose that Σ ∪ {ϕ} is satisfiable. Then there is a truth assignmentv satisfying Σ and ϕ; by definition (Proposition 0.2.3), v does not satisfy ¬ϕ.This means that Σ does not entail ¬ϕ.
Conversely suppose that Σ 6|= ¬ϕ. There is therefore a truth assignement vsatisfying Σ, but not satisfying ¬ϕ. It follows v(ϕ) = T , so v actually satisfiesΣ ∪ {ϕ}.
Theorem 0.2.11. Let Σ0 be a finite theory and ϕ a wff. Then there is analgorithm which decides whether Σ entails ϕ or not.
Proof . In Σ0 ∪ {ϕ}, only finitely many sentence symbols occur. List all (butfinitely many) truth tables involving them. If all truth tables satisfying Σ0 alsosatisfy ϕ, the answer is “yes”; otherwise answer “no” at first flaw.
Question. And what about an infinite set?
This question will find an answer when we can reduce the infinite to the finite(Compactness, Theorem 0.5.2). The answer is exactly half as good (Corollary0.5.14). This is far from obvious, as Σ might now involve infinitely many sen-tence symbols, giving rise to continuum many truth tables. However ϕ is asingle formula, so one has the feeling that some finite subset Σ0 of Σ wouldsuffice to entail ϕ, in case Σ does.
There is however no bound a priori on how many sentence symbols areinvolved in Σ0, as shows the example ϕ = A1, Σ = {A2, A2,→ A3, . . . , An−1 →An, An → A1}. This is actually the reason why the algorithm for infinite Σgiven in Corollary 0.5.14 cannot be implemented to answer “no”: if it hasn’tanswered “yes” after a long waiting time, it could be
• either because Σ |= ϕ, but we haven’t waited long enough to be positiveabout it;
• or simply because Σ 6|= ϕ, but we’ll never know.
These questions will be made more precise in §0.5.2.
0.2.3 Simplifying the LanguageFive connectives are too much: they made inductive definitions and proofsextremely clumsy. Moreover we have a feeling that they are redundant. Weshall cut down on symbols and rules.
In Proposition 0.2.3, we have treated ∧,∨,↔ as if they were elementaryconnectives, and not abbreviations. We show that considering them as abbre-viations is harmless. This involves rewriting the whole theory with only twoconnectives, and checking that this new setting is equivalent to the previousone.
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Notation 0.2.12. The collection WFF′ is the smallest set such that:
• each An is a wff;
• if ϕ and ψ are wff, so are ¬ϕ and ϕ→ ψ.
The set WFF′ would require a version of the Unique Readability Theorem(Theorem 0.1.9), which may be regarded as a special case of the previous one,since WFF′ ⊆WFF. In any case, we keep working with few parentheses.
There is of course a natural way to translate a “standard” wff ϕ (using fiveconnectives) into a wff’ ϕ′ using only ¬ and →.
Notation 0.2.13. We inductively translate any wff ϕ into a formula ϕ′ whichuses only the two connectives ¬ and →:
• A′n is An for each n ∈ N;
• (¬ϕ)′ is (¬ϕ′);
• (ϕ ∧ ψ)′ is ¬(ϕ′ → ¬ψ′);
• (ϕ ∨ ψ)′ is (¬ϕ′ → ψ′);
• (ϕ→ ψ)′ is (ϕ′ → ψ′);
• (ϕ↔ ψ)′ is ¬((ϕ′ → ψ′)→ ¬(ψ′ → ϕ′)).
Hence for any formula ϕ, the translation ϕ′ uses only two connectives, andintuitively bears the same meaning.
Example 0.2.14. Let ϕ be (A1 ↔ A2)∧(A3∨A1). Then ϕ′ is ¬((A1 ↔ A2)′ →¬(A3 ∨A1)′), which is ¬(¬((A1 → A2)→ ¬(A2 → A1))→ ¬(¬A3 → A1)).
The issue is that this new notion of formula gives rise to another extensionof truth assignment, and therefore to a new notion of entailment. We mustcheck that these new notions do coincide with their original “five connectives”analogues.
Notation 0.2.15. For a truth assignment v : WFF → {F, T}, let v′ be theextension of v defined by:
• v′(An) = v(An) for each n ∈ N;
• v′(¬ψ) = T iff v(ψ) = F ;
• v′(ψ1 → ψ2) = T iff v(ψ1) = F or v(ψ2) = T .
Notice that this is a subset of the conditions defining v (Proposition 0.2.3),and that v′(ψ) is defined only for ψ ∈WFF′.
Lemma 0.2.16. Let v be a truth assignment, ϕ a wff, and ϕ′ its translationusing only ¬ and →. Then v(ϕ) = v′(ϕ′).
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Proof . By induction on ϕ.
• If ϕ is a sentence symbol, then by definition v(ϕ) = v(ϕ) = v′(ϕ′).
• Suppose that ϕ is ¬ψ. Then v(ϕ) = T is equivalent to v(ψ) = F , which isby induction equivalent to v′(ψ′) = F , which is equivalent to v′(ϕ′) = T .
• Suppose that ϕ is ψ1 → ψ2. Then v(ϕ) = T iff v(ψ1) = F or v(ψ2) = Tiff v′(ψ′1) = F or v′(ψ′2) = T iff v′(ψ′1 → ψ′2) = T iff v′(ϕ′) = T .
• Suppose that ϕ is ψ1 ∧ ψ2. Then v(ϕ) = T iff v(ψ1) = v(ψ2) = T iffv′(ψ′1) = v′(ψ′2) = T iff v′(¬(ψ′1 → ¬ψ′2)) = T iff v′(ϕ′) = T .
• We proceed similarly for the remaining two connectives ∨ and ↔.
Any modification, even apparently harmless, of the notion of truth assign-ment, induces a modfication of the notion of semantic consequence (Definition0.2.9).
Notation 0.2.17. Let Θ be a consistent subset of WFF′ and ψ ∈WFF′. WriteΘ |=′ ψ if whenever v′ satisfies Θ, then v′(ψ) = T .
Corollary 0.2.18. Σ |= ϕ iff Σ′ |=′ ϕ′.
Proof . Suppose that Σ |= ϕ. We show that Σ′ |=′ ϕ′. Let v be a truthassignement such that v′ satisfies Σ′. By Lemma 0.2.16, v satisfies Σ. Byassumption, v satisfies ϕ. Using Lemma 0.2.16 again, v′ satisfies ϕ′.
Now suppose that Σ′ |=′ ϕ′. We show that Σ |= ϕ. Let v be a truth assign-ment such that v satisfies Σ. By Lemma 0.2.16, v′ satisfies Σ′. By assumption,v′ satisfies ϕ′. Using Lemma 0.2.16 again, v satisfies ϕ.
From now on, we know that the semantics induced by two connectives is thesame as that induced by five. In particular we happily forget about v′ and |=′.(We are not entirely done with ϕ′ however, since a similar verification will haveto be carried on the syntactic side, §0.3.3.)
We shall also identify any truth assignment v with its extension v.
End of Lecture 2.
Lecture 3 (Natural Deduction in Classical Logic)
0.3 Natural Deduction (in Classical Logic)In this section we formalize the notions of deduction and syntactic consequence;this will be done in §0.3.1. We shall work in the framework of Natural De-duction (a certain notion of deduction), using Classical Logic (certain rules ofdeduction).
Natural Deduction was invented by Gerhard Gentzen; there are other notionsof deduction (for instance, Hilbert’s system, which relies on modus ponens), and
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there are other logics (for instance, intuitionistic logic, which does not allowcontradiction proofs). They all yield different notions of deduction. We madethe choices which best reflect a normal mathematician’s thought.
After a couple of examples (§0.3.2), we shall proceed in §0.3.3 to reducingthe language to two connectives. There are indeed redundancies as we haveobserved (Corollary 0.2.18). We shall show that one may restrict the languageand deduction rules to ¬ and → without affecting the power of the notion ofdeduction.
0.3.1 Syntactic Consequences and DeductionsWhat is a consequence, again? In terms of proofs, ϕ follows from assumptionsΣ if there is a way to deduce ϕ using the wff’s of the theory Σ.
Definition 0.3.1 (deduction). A deduction is a finite sequence of basic stepswhich are regarded as elementary deduction rules.
The deduction can be drawn as a tree showing the consecutive steps. Thiswill be more precise when we have given an example of a set of such rules, butwe can already formulate the following important definition.
Definition 0.3.2 (syntactic consequence). Let Σ be a theory and ϕ a wff. ϕ isa theorem of Σ, denoted Σ ` ϕ, if there is a deduction of ϕ under Σ. One alsosays that Σ proves ϕ.
If Σ = ∅, one writes ` ϕ, and says that ϕ is a theorem (of our logic).
Example 0.3.3. At the end of the day, (ϕ∧(ϕ→ ψ))→ ψ should be a theorem.
We now explain the basic steps of deduction. Bear in mind that we chosethe most natural setting (natural deduction), but there are many other possiblesettings, and not all are equivalent. A characteristic feature of natural deductionis that connectives may be introduced and eliminated. Describing the deductionrules amounts to explaining how connectives are “proved and used”. In whatfollows, Σ stands for a (possibly empty) theory, and ϕ,ψ, etc. for wff’s.
• There is only one axiom (so far). In natural deduction for propositionallogic, the only way to start a proof is by assuming something.
• Weakening allows us to add unused assumptions to an existing deduction.
• The case of negation is very subtle. We have a way to introduce a nega-tion provided we deduced an inconsistency (¬i), but we also have a wayof removing double-negations (¬e). This is a characteristic feature of clas-sical logic, which is actually equivalent to admitting reductio ad absur-dum. Keep in mind that this is only one among the possible treatments of
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negation and contradiction, though it seems satisfactory to us as workingmathematicians.The elimination rule ¬e is necessary to show the excluded middle law(Theorem 0.3.11 below); if we introduce a weaker rule, excluded middleneed not hold any more. Exercise 0.10
How hard it may be to believe at first, ¬i is not related to reduction toabsurdum. Only ¬e is.
• There are two elimination rules for ∧ as there are two possible ways touse a conjunction.
• Dually, there are two introduction rules for ∨, as there are two ways toprove a disjunction. This is in general difficult, since when trying to showϕ∨ψ, one never knows a priori which is true; contradiction proofs will behelpful (see the proof of Theorem 0.3.11 below).On the other hand ∨e may look subtle at first sight, but it merely describesthe process of a case-division.
• ∧ and ∨ may be regarded as convenient abbreviations (this will be madeformal in §0.3.3). But the connective → is essential.
The elimination rule →e is classically called modus ponens; dually theintroduction rule →i is called modus tollens.
• It is clear that ↔ should have the moral status of an abbreviation. Yet incase it is admitted as a primitive connective, we need deduction rules.
We shall start using quantifiers in the next chapters and introduce adequaterules; for the moment they won’t appear, as we work in Propositional Logic.
Remark 0.3.4. Notice that in our deductions we may have several assumptions,but at each step we have only one conclusion. This is another typical feature ofNatural Deduction which supposedly reflects the way we mathematicians work.But there are other, more complex, deduction systems (for instance, Gentzen’sSequent Calculus).
A formalization of deduction yields the following essential notion.
Definition 0.3.5 (consistency). A theory Σ is consistent if there is no wff ϕsuch that Σ ` ϕ and Σ ` ¬ϕ.
17
In other words, a theory is consistent if it is not possible to deduce a contra-diction from it. There is a symbol for inconsistency (or contradiction), whichwe shall not use. Exercise 0.9
Lemma 0.3.6. Σ is inconsistent iff Σ ` ϕ for all wff ϕ.
Proof . Suppose Σ inconsistent. Then there is ψ such that Σ ` ψ and Σ ` ¬ψ.Hence
Σ ` ψΣ ∪ {¬ϕ} ` ψ
WkΣ ` ¬ψ
Σ ∪ {¬ϕ} ` ¬ψWk
Σ ` ¬¬ϕΣ ` ϕ
¬e
¬i
The converse is obvious.
0.3.2 Our First DeductionsWe begin with an easy example.
Example 0.3.7. For wff’s ϕ, ψ, one has ` (ϕ↔ ψ)↔ ((ϕ→ ψ) ∧ (ψ → ϕ))
Verification: On the one hand,
ϕ↔ ψ ` ϕ↔ ψAx
ϕ↔ ψ ` ϕ→ ψ↔e
ϕ↔ ψ ` ϕ↔ ψAx
ϕ↔ ψ ` ψ → ϕ↔e
ϕ↔ ψ ` (ϕ→ ψ) ∧ (ψ → ϕ)` (ϕ↔ ψ)→ ((ϕ→ ψ) ∧ (ψ → ϕ))
→i
∧i
On the other hand,
(ϕ→ ψ) ∧ (ψ → ϕ) ` ((ϕ→ ψ) ∧ (ψ → ϕ))Ax
(ϕ→ ψ) ∧ (ψ → ϕ) ` ϕ→ ψ∧e
(∗1)Ax
(∗2)∧e
((ϕ→ ψ) ∧ (ψ → ϕ)) ` ϕ↔ ψ
` ((ϕ→ ψ) ∧ (ψ → ϕ))→ (ϕ↔ ψ)→i
↔i
where (∗1) and (∗2) are the symmetric deductions obtained by exchanging ϕand ψ.
Combining both trees with ↔i, we find
` (ϕ↔ ψ)↔ ((ϕ→ ψ) ∧ (ψ → ϕ)) ♦
Theorem 0.3.8 (Contraposition). Σ ` ϕ→ ψ iff Σ ` ¬ψ → ¬ϕ.
Proof . Suppose Σ ` ϕ→ ψ. Then
Σ ` ϕ→ ψ
Σ ∪ {¬ψ,ϕ} ` ϕ→ ψWk
{ϕ} ` ϕAx
Σ ∪ {¬ψ,ϕ} ` ϕWk
Σ ∪ {¬ψ,ϕ} ` ψ→e
{¬ψ} ` ¬ψAx
Σ ∪ {¬ψ,ϕ} ` ¬ψWk
Σ ∪ {¬ψ} ` ¬ϕ¬i
Σ ` ¬ψ → ¬ϕ→i
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Suppose Σ ` ¬ψ → ¬ϕ. What we just proved implies Σ ` ¬¬ϕ → ¬¬ψ.Therefore
{ϕ} ` ϕAx
{ϕ,¬ϕ} ` ϕWk
{¬ϕ} ` ¬ϕAx
{ϕ,¬ϕ} ` ¬ϕWk
{ϕ} ` ¬¬ϕΣ ∪ {ϕ} ` ¬¬ϕ
Wk
¬iΣ ` ¬¬ϕ→ ¬¬ψ
Σ ∪ {ϕ} ` ¬¬ϕ→ ¬¬ψWK
Σ ∪ {ϕ} ` ¬¬ψΣ ∪ {ϕ} ` ψΣ ` ϕ→ ψ
→i
¬e
→e
Parallel to the relationship between entailment and satisfiability (Lemma0.2.10) runs the following.
Lemma 0.3.9 (see Lemma 0.2.10). Σ ∪ {ϕ} is not inconsistent iff Σ 6` ¬ϕ.
Proof . Suppose Σ ∪ {ϕ} is not inconsistent and Σ ` ¬ϕ. Then Σ ∪ {ϕ} ` ϕ(by axiom and weakening), and Σ ∪ {ϕ} ` ¬ϕ (by assumption). It follows thatΣ ∪ {ϕ} is inconsistent, against the assumption. So Σ 6` ¬ϕ.
Suppose that Σ 6` ¬ϕ, and Σ ∪ {ϕ} is inconsistent. By inconsistency , thereis ψ be such that Σ ∪ {ϕ} ` ψ and Σ ∪ {ϕ} ` ¬ψ. Then by ¬i, Σ ` ¬ϕ, againstthe assumption. So Σ ∪ {ϕ} is not inconsistent.
Notice that the latter proof did not involve a semantic analog of the elimi-nation of double negation (¬e), but only ¬i; whence the odd statement. As youbelieve in ¬e, you may say “is consistent” for “is not inconsistent”.
Theorem 0.3.10 (Reductio ad Absurdum). If Σ ∪ {¬ϕ} is inconsistent thenΣ ` ϕ.
Proof . Suppose Σ ∪ {¬ϕ} inconsistent. By Lemma 0.3.9, Σ ` ¬¬ϕ. By (¬e),Σ ` ϕ.
Still not impressed?
Theorem 0.3.11 (Excluded Middle holds in Classical Logic). Excluded middleis a theorem of Classical Logic.
That is: for any wff ϕ, one has ` ϕ ∨ ¬ϕ. Notice that one does not know apriori which of ϕ or ¬ϕ to prove, so there is little chance to find a quick deductionrelying entirely on ∨i. We shall work our way through a contradiction proofinvolving ¬e.
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Proof . Due to pagesetting limitations, I must pause and resume.
{ϕ} ` ϕAx
{¬(ϕ ∨ ¬ϕ), ϕ} ` ϕWk
{¬(ϕ ∨ ¬ϕ), ϕ} ` ϕ ∨ ¬ϕ∨i
{¬(ϕ ∨ ¬ϕ)} ` ¬(ϕ ∨ ¬ϕ)Ax
{¬(ϕ ∨ ¬ϕ), ϕ} ` ¬(ϕ ∨ ¬ϕ)Wk
¬(ϕ ∨ ¬ϕ) ` ¬ϕ¬(ϕ ∨ ¬ϕ) ` ϕ ∨ ¬ϕ
∨i︸ ︷︷ ︸(∗)
(∗)¬(ϕ ∨ ¬ϕ) ` ϕ ∨ ¬ϕ ¬(ϕ ∨ ¬ϕ) ` ¬(ϕ(∨¬ϕ)
Ax
` ¬¬(ϕ ∨ ¬ϕ)` ϕ ∨ ¬ϕ
¬e
¬i
The proof of Theorem 0.3.11 shows that contradiction deductions (typicallyexcluded middle) are always a little counter-intuitive to formalize. Notice thatin some weaker deduction systems, excluded middle cannot be deduced!
End of Lecture 3.
Lecture 4 (Reduction to Two Connectives (1/2))
0.3.3 Simplifying the Language (and Presentation)We have introduced a battery of deduction rules, though we had concluded in§0.2.3 that two connectives were enough to convey all the desired meaning. Wenow must check that working with only two connectives would preserve thedeductive strength, i.e. check that all deductions can be imitated even whenone restricts connectives and rules to the (¬,→) setting.
Recall what ϕ′ stands for (Notation 0.2.13). It is intuitively clear that prov-ing ϕ and proving ϕ′ should be the same.
Lemma 0.3.12 (see Lemma 0.2.16). {ϕ} ` ϕ′ and {ϕ′} ` ϕ.
Proof . By induction on ϕ. Provided the result has been proved for a formulaψ, we shall freely write:
Σ ` ψΣ ` ψ′
induction
though induction is of course not a deduction rule. This is actually short-handfor the following:
Σ ` ψ
.... induction
{ψ} ` ψ′
` ψ → ψ′→i
Σ ` ψ → ψ′Wk
Σ ` ψ′→e
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Of courseΣ ` ψ′Σ ` ψ induction
is a similar shortcut, and is licit for a similar reason.
• Suppose that ϕ is a sentence symbol. Then ϕ′ = ϕ, and the claim isobvious.
• Suppose that ϕ is ¬ψ. Then ϕ′ = ¬(ψ′). Now
{ϕ} ` ϕAx
{ϕ,ψ′} ` ϕWk
{ψ′} ` ψ′Ax
{ϕ,ψ′} ` ψ′Wk
{ϕ,ψ′} ` ψinduction
{ϕ} ` ¬ψ′︸︷︷︸=ϕ′
¬i
and
{ϕ′} ` ϕ′Ax
{ϕ′, ψ} ` ϕ′Wk
{ψ} ` ψAx
{ϕ′, ψ} ` ψWk
{ϕ′, ψ} ` ψ′induction
{ϕ} ` ¬ψ︸︷︷︸=ϕ′
¬i
• Suppose that ϕ is (ψ1 → ψ2). Then
{ϕ} ` ϕAx
{ϕ,ψ′1} ` ϕWk
{ψ′1} ` ψ′1Ax
{ϕ,ψ′1} ` ψ′1Wk
{ϕ,ψ′1} ` ψ1induction
{ϕ,ψ′1} ` ψ2
{ϕ,ψ′1} ` ψ′2{ϕ} ` ϕ′
→i
induction
→e
On the other hand
{ϕ′} ` ϕ′Ax
{ϕ′, ψ1} ` ϕ′Wk
{ψ1} ` ψ1Ax
{ϕ′, ψ1} ` ψ1Wk
{ϕ′, ψ1} ` ψ′1induction
{ϕ′, ψ1} ` ψ′2{ϕ′, ψ1} ` ψ2
{ϕ′} ` ϕ→i
induction
→e
Of course the case of ¬ and → was trivial. The real proof starts here.
• Suppose that ϕ = ψ1 ∧ ψ2, so that ϕ′ = (¬(ψ′1 → ¬ψ′2)).
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– One has
{ϕ} ` ϕAx
{ϕ,ψ′1 → ¬ψ′2} ` ϕWk
{ϕ,ψ′1 → ¬ψ′2} ` ψ1
{ϕ,ψ′1 → ¬ψ′2} ` ψ′1induction
∧e{ψ′1 → ¬ψ′2} ` ψ′1 → ¬ψ′2
Ax
{ϕ,ψ′1 → ¬ψ′2} ` ψ′1 → ¬ψ′2Wk
{ϕ,ψ′1 → ¬ψ′2} ` ¬ψ′2→e︸ ︷︷ ︸
(∗)
Hence
(∗){ϕ,ψ′1 → ¬ψ′2} ` ¬ψ′2
{ϕ} ` ϕAx
{ϕ,ψ′1 → ¬ψ′2} ` ϕWk
{ϕ,ψ′1 → ¬ψ′2} ` ψ2
{ϕ,ψ′1 → ¬ψ′2} ` ψ′2induction
∧e
{ϕ} ` ¬(ψ′1 → ¬ψ′2)¬i
– On the other hand
{¬ψ1} ` ¬ψ1Ax
{ϕ′,¬ψ1, ψ′1, ψ′2} ` ¬ψ1
Wk
{ψ′1} ` ψ′1Ax
{ϕ′,¬ψ1, ψ′1, ψ′2} ` ψ′1
Wk
{ϕ′,¬ψ1, ψ′1, ψ′2} ` ψ1
induction
{ϕ′,¬ψ1, ψ′1} ` ¬ψ′2
¬i
{ϕ′,¬ψ1} ` ψ′1 → ¬ψ′2→i︸ ︷︷ ︸
(†)
Therefore
(†){ϕ′,¬ψ1} ` ψ′1 → ¬ψ2
{ϕ′} ` ¬(ψ′1 → ¬ψ′2)Ax
{ϕ′,¬ψ1} ` ¬(ψ′1 → ¬ψ′2)Wk
{ϕ′} ` ¬¬ψ1¬i
{ϕ′} ` ψ1¬e
Moreover
{¬ψ′2} ` ¬ψ′2Ax
{ϕ′, ψ′1,¬ψ′2} ` ¬ψ′2Wk
{ϕ′,¬ψ′2} ` ψ′1 → ¬ψ′2→i
{ϕ′} ` ϕ′Ax
{ϕ′,¬ψ′2} ` ϕ′Wk
{ϕ′} ` ¬¬ψ′2¬i
{ϕ′} ` ψ′2¬e
{ϕ′} ` ψ2induction
Combining both threes, we find {ϕ′} ` ϕ.
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• Suppose that ϕ = ψ1 ∨ ψ2, so that ϕ′ = (¬ψ′1 → ψ′2).
– {ψ1,¬ψ′1} is clearly inconsistent by induction. By Lemma 0.3.6,{ψ1,¬ψ′1} ` ψ′2 and {ψ1} ` ϕ′.On the other hand, {ψ2,¬ψ′1} ` ψ′2 by induction, so {ψ2} ` ϕ′.As ϕ = ψ1 ∨ ψ2, combining both deduction trees and eliminating ∨,we find {ϕ} ` ϕ′. This was the easy part.
– For the converse we need an easy remark:
{ψ′1} ` ψ′1Ax
{ψ′1} ` ψ1induction
{¬ϕ,ψ′1} ` ψ1Wk
{¬ϕ,ψ′1} ` ψ1 ∨ ψ2∨i
{¬ϕ} ` ¬ϕAx
{¬ϕ,ψ′1} ` ¬ϕWk
{¬ϕ} ` ¬ψ′1¬i︸ ︷︷ ︸
(∗1)
Of course one has similarly {¬ϕ} ` ¬ψ′2 (∗2).Therefore
(∗1){¬ϕ} ` ¬ψ′1{ϕ′,¬ϕ} ` ¬ψ′1
Wk{ϕ′} ` ¬ψ′1 → ψ′2
Ax
{ϕ′,¬ϕ} ` ¬ψ′1 → ψ′2Wk
{ϕ′,¬ϕ} ` ψ′2→e
(∗2){¬ϕ} ` ¬ψ′2{ϕ′,¬ϕ} ` ¬ψ′2
Wk
{ϕ′} ` ¬¬ϕ¬i
{ϕ′} ` ϕ¬e
• The case of ↔ is left as an exercise, but essentially easier than ∨.
Hence ϕ and its translation ϕ′ prove each other. Of course they do so onlyin the setting involving all connectives (since ϕ may involve ∧); Lemma 0.3.12is but a first step as the notion of deduction ` itself has changed.
Notation 0.3.13. Let Σ be a theory and ϕ a wff. Write Σ `′ ϕ if there isa deduction of ϕ from Σ using only axiom, weakening and the deduction rulesassociated to ¬ and →.
Preservation of the notion of deduction under translation is therefore thefollowing result.
Corollary 0.3.14 (see Corollary 0.2.18). Let Σ be a theory and ϕ a wff. ThenΣ ` ϕ iff Σ′ `′ ϕ′.
Proof . For the moment we prove only that Σ′ `′ ϕ′ implies Σ ` ϕ. Sup-pose Σ′ `′ ϕ′. By finiteness of the notion of a deduction, there are formulasψ1, . . . , ψn ∈ Σ such that {ψ′1, . . . , ψ′n} `′ ϕ′. Using →i repeatedly, one finds
`′ ψ′1 → (ψ′2 → . . . (ψ′n → ϕ′). . .)
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As all the rules in `′ are in `, one even has
` ψ′1 → (ψ′2 → . . . (ψ′n → ϕ′). . .)
Now one sees that
(ψ1 → (ψ2 → . . . (ψn → ϕ). . .))′ is ψ′1 → (ψ′2 → . . . (ψ′n → ϕ′). . .)
So by Lemma 0.3.12, it follows
` ψ1 → (ψ2 → . . . (ψn → ϕ). . .)
In particular, {ψ1, . . . , ψn} ` ϕ, and Σ ` ϕ.We shall prove the converse implication next time.
End of Lecture 4.
Lecture 5 (Reduction to Two Connectives (2/2); Soundness)We finish the proof of Corollary 0.3.14.
Proof of Corollary 0.3.14, completed. The remaining proof that Σ ` ϕ im-plies Σ′ `′ ϕ′ is by induction on the length of the deduction.
• The cases of axiom and weakening are clear.
• Suppose the last step is for instance ¬i:
Σ ∪ {ψ} ` χ Σ ∪ {ψ} ` ¬χΣ ` ¬ψ
¬i
By induction, one has Σ′ ∪ {ψ′} `′ χ′ and Σ′ ∪ {ψ′} `′ ¬χ′. As ¬i is in`′, it follows
Σ′ ∪ {ψ′} `′ χ′ Σ′ ∪ {ψ′} `′ ¬χ′
Σ′ `′ ¬ψ′¬i
And therefore Σ′ `′ ¬ψ′.
• It is clear that the cases of ¬e,→i, and→e are as trivial (since these rulesare in `′). We now turn to the interesting configurations.
• Suppose the last step is ∧i, that isΣ ` ψ1 Σ ` ψ2
Σ ` ψ1 ∧ ψ2∧i
We aim at proving Σ′ `′ (ψ1∧ψ2)′; recall that (ψ1∧ψ2)′ is ¬(ψ′1 → ¬ψ′2).By induction, we know Σ′ `′ ψ′1 and Σ′ ` ψ′2. Therefore
.... induction
Σ′ `′ ψ′1Σ′ ∪ {ψ′1 → ¬ψ′2} `′ ψ′1
{ψ′1 → ¬ψ′2} `′ ψ′1 → ¬ψ′2Ax
Σ′ ∪ {ψ′1 → ¬ψ′2} `′ ψ′1 → ¬ψ′2Wk
Σ′ ∪ {ψ′1 → ¬ψ′2} `′ ¬ψ′2→e︸ ︷︷ ︸
(∗)
24
and
(∗)Σ′ ∪ {ψ′1 → ¬ψ′2} `′ ¬ψ′2
.... induction
Σ′ `′ ψ′2Σ′ ∪ {ψ′1 → ¬ψ′2} `′ ψ′2
Wk
Σ′ `′ ¬(ψ′1 → ¬ψ′2)¬i
• Suppose the last step is ∧e1 :
Σ ` ψ1 ∧ ψ2Σ ` ψ1
∧e1
By induction, Σ′ `′ ¬(ψ′1 → ¬ψ′2). Clearly Σ′ ∪ {¬ψ′1} `′ ψ′1 → ¬ψ′2 (∗).Hence
(∗)Σ′ ∪ {¬ψ′1} `′ ψ′1 → ¬ψ′2
.... induction
Σ′ `′ ¬(ψ′1 → ¬ψ′2)Σ′ ∪ {¬ψ′1} `′ ¬(ψ′1 → ¬ψ′2)
Wk
Σ′ `′ ¬¬ψ′1¬i
Σ′ `′ ψ′1¬e
Suppose on the other hand that the last step is ∧e2 :
Σ ` ψ1 ∧ ψ2Σ ` ψ2
∧e2
By induction, Σ′ `′ ¬(ψ′1 → ¬ψ′2). Of course Σ′∪{¬ψ′2} `′ ψ′1 → ¬ψ′2 (†).Thus
(†)Σ′ ∪ {¬ψ′2} `′ ψ′1 → ¬ψ′2
.... induction
Σ′ `′ ¬(ψ′1 → ¬ψ′2)Σ′ ∪ {¬ψ′2} `′ ¬(ψ′1 → ¬ψ′2)
Wk
Σ′ `′ ¬¬ψ′2¬i
Σ′ `′ ψ′2¬e
• Suppose that the last step is ∨i1 :
Σ ` ψ1Σ ` ψ1 ∨ ψ2
∨i1
By induction, we know Σ′ `′ ψ′1. From Lemma 0.3.9 we know that Σ′ ∪{¬ψ′1} is inconsistent, so by Lemma 0.3.6 Σ′ ∪ {¬ψ′1} `′ ψ′2. Introducing→, we find Σ′ `′ ¬ψ′1 → ψ′2, that is Σ′ `′ (ψ1 ∨ ψ2)′.∨i2 is dealt with similarly.
25
• Suppose the last step of the form:Σ ` ψ1 ∨ ψ2 Σ ∪ {ψ1} ` χ Σ ∪ {ψ2} ` χ
Σ ` χ∨e
By induction, Σ′ `′ ¬ψ′1 → ψ′2, Σ′ ∪ {ψ′1} `′ χ′, and Σ′ ∪ {ψ′2} `′ χ′. Bycontraposition (Theorem 0.3.8, which still holds for `′), we easily find:
Σ′ ∪ {¬χ′} `′ ¬ψ′1 (∗1) and Σ′ ∪ {¬χ′} `′ ¬ψ′2 (∗2)
Therefore.... induction
Σ′ `′ ¬ψ′1 → ψ′2Σ′ ∪ {¬χ′} `′ ¬ψ′1 → ψ′2
(∗1)Σ′ ∪ {¬χ′} `′ ¬ψ′1
Σ′ ∪ {¬χ′} `′ ψ′2→e
(∗2)Σ′ ∪ {¬χ′} `′ ¬ψ′2
Σ′ `′ ¬¬χ′¬i
Σ′ `′ χ′¬e
• The ↔ rules are easy to deal with.All is well that ends well and we may forget about `′ and ϕ′. From now on
there are two connectives: ¬ and →. The other three are abbreviations. Have Isaid that before? It is very important to say it again, and above all, it is crucialthat a redundant connective should be regarded a an abbreviation of the sameexpression in either setting (syntactic or semantic).
However we shall be extremely hypocritical about connectives. In our defi-nitions and inductive proofs, we need only to consider ¬ and →, but whenevernecessary we shall feel free to use ∧, ∨, and ↔ and the related rules. We knowthat these are harmless shortcuts, which preserve the notions of truth and ofdeduction.
We now aim at simplifying the presentation of deductions.Notation 0.3.15. When we write a deduction under a set of assumptions Σ, wewrite Σ on top (“deduction under Σ”). Whenever an assumption ϕ is eliminated(via ¬i or →i), we cross it out.
Let us give an example. Suppose we want to show Σ ` ϕ→ ψ. Instead ofΣ ∪ {ϕ} ` . . .
...Σ ∪ {ϕ} ` ψΣ ` ϕ→ ψ
→i
we now write:Σ ϕ
...ψ
and thenΣ 6 ϕ
...ψ
ϕ→ ψ
26
As Σ remains uncrossed, one has Σ ` ϕ→ ψ.Similarly,
Σ ∪ {ϕ} ` ψ Σ ∪ {ϕ} ` ¬ψΣ ` ¬ϕ
¬i
becomes:Σ ϕ
ψ
Σ ϕ
¬ψthen Σ 6 ϕ
ψ
Σ 6 ϕ¬ψ
¬ϕ
which means Σ ` ¬ϕ. Do not forget to cross out every relevant occurence of ϕ!
Example 0.3.16. As an example, we prove Excluded Middle (Theorem 0.3.11)with this notation.
¬(ϕ ∨ ¬ϕ) ϕ1.
¬(ϕ ∨ ¬ϕ) ϕϕ ∨ ¬ϕ ∨i
2.
¬(ϕ ∨ ¬ϕ) 6 ϕϕ ∨ ¬ϕ¬ϕ ¬i
∨i
3.
¬(ϕ ∨ ¬ϕ) 6 ϕϕ ∨ ¬ϕ¬ϕ
ϕ ∨ ¬ϕ ∨i
¬i
∨i
4.
¬(ϕ 6 ∨¬ϕ) 6 ϕϕ ∨ ¬ϕ¬ϕ
ϕ ∨ ¬ϕ¬¬(ϕ ∨ ¬ϕ)
¬i
∨i
¬i
∨i
5.
¬(ϕ 6 ∨¬ϕ) 6 ϕϕ ∨ ¬ϕ¬ϕ
ϕ ∨ ¬ϕ¬¬(ϕ ∨ ¬ϕ)ϕ ∨ ¬ϕ ¬e
¬i
∨i
¬i
∨i
6.
0.3.4 The Soundness TheoremRecall that we have two notions of consequence: semantic (|=, Definition 0.2.9)and syntactic (`, Definition 0.3.2). Once we have proved the Soundness andCompleteness Theorems (Theorems 0.3.17 and 0.4.1), all will be well that endswell: they do coincide. The easier part comes first.
Theorem 0.3.17 (soundness of propositional logic). Let Σ be a theory and ϕbe a wff. If Σ ` ϕ, then Σ |= ϕ.
Proof . Induction on the length of the deduction. We use Corollaries 0.2.18 and0.3.14, which enable us to treat only the rules associated to ¬ and →.
• If the deduction is just an axiom, it is of the form
{ϕ} ` ϕAx
Clearly {ϕ} |= ϕ.
27
• The case of weakening is absolutely trivial.
• Suppose that the last step is by ¬i; that is, Σ ` ¬ϕ because Σ ∪ {ϕ}is inconsistent. By induction, there are no truth assignments satisfyingΣ∪ {ϕ}. So any truth assignment satisfying Σ (if any) satisfies ¬ϕ. ThusΣ |= ¬ϕ (notice that we did not claim satisfiability of Σ).
• Suppose that the last step is
Σ ` ¬¬ϕΣ ` ϕ
¬e
By induction, we know that Σ |= ¬¬ϕ. Let v be a truth assignmentsatisfying Σ. Then v(ϕ) = v(¬¬ϕ) = T , we are done.
• Suppose that the last step of the deduction is the modus ponens:
Σ ` ϕ→ ψ Σ ` ϕΣ ` ψ
→e
Hence Σ ` ϕ→ ψ and Σ ` ϕ hold; by induction, Σ |= ϕ→ ψ and Σ |= ϕ.Let v be a truth assignment satisfying Σ. Then v(ϕ→ ψ) = v(ϕ) = T , sov(ψ) = T . This shows Σ |= ψ.
• Suppose that the last step of the deduction is the modus tollens:
Σ ∪ {ϕ} ` ψΣ ` ϕ→ ψ
→i
By induction, Σ ∪ {ϕ} |= ψ; we aim at showing Σ |= ϕ → ψ. Let v bea truth assignment satisfying Σ. If v does not satisfy ϕ then v satisfiesϕ→ ψ by definition. If v does satisfy ϕ then v satisfies Σ ∪ {ϕ}; we thenknow that v satisfies ψ, and therefore v satisfies ϕ→ ψ again. In any casev satisfies ϕ→ ψ. This implies that Σ |= ϕ→ ψ.
Even with five connectives the case of ∧ etc. wouldn’t be very complicated.Induction completes the proof. Exercise 0.11
Exercise 0.12
End of Lecture 5.
Lecture 6 (Completeness of Propositional Logic)
0.4 The Completeness TheoremTheorem 0.3.17 states that every syntactic consequence is a semantic conse-quence; roughly speaking, “if you can prove it, it is true”. Very remarkably(and fortunately), the converse holds.
28
Theorem 0.4.1 (completeness of propositional logic). Let Σ be a theory andϕ be a wff. If Σ |= ϕ, then Σ ` ϕ.
Lemma 0.4.2. The Completeness Theorem is equivalent to:
every consistent theory is satisfiable.
Proof . Suppose the Completeness Theorem. Let Σ be a consistent theory.As Σ is consistent (Definition 0.3.5), there is a wff ϕ such that Σ 6` ϕ. Bycompleteness, this implies Σ 6|= ϕ. It follows from Lemma 0.2.10 that Σ ∪ {ϕ}is satisfiable; let v be a truth assignment satisfying Σ ∪ {ϕ}. In particular, vsatisfies Σ!
Suppose that every consistent theory is satisfiable. Let Σ 6` ϕ. Then Σ∪{¬ϕ}is consistent by Lemma 0.3.9. In particular, Σ ∪ {¬ϕ} is satisfiable. So there isa truth assignment v satisfying Σ ∪ {¬ϕ}: v satisfies Σ but does not satisfy ϕ,so Σ 6|= ϕ.
0.4.1 Extending the TheoryWe shall prove the Completeness Theorem. In view of Lemma 0.4.2, it sufficesto show that every consistent theory is satisfiable. So, given a consistent set,we must find a truth assignment satisfying it.
As often in mathematics, choice is embarrassing. We shall force ourself tohave a unique suitable truth assignment. This amounts to maximizing, in aconsistent way, the set of requirements.
Definition 0.4.3 (complete). A consistent theory Σ is complete if for any wffϕ, either ϕ or ¬ϕ is in Σ.
The following remark will be implicitely used in the proof.
Remark 0.4.4. If Σ is complete, then Σ ` ϕ iff ϕ ∈ Σ.
Verification: Suppose Σ ` ϕ, but ϕ 6∈ Σ. By completeness, ¬ϕ ∈ Σ. So Σ ` ¬ϕand Σ ` ϕ, against consistency. ♦
We shall extend any consistent theory to a complete one. As we have said,the problem of satisfying a complete consistent set is actually easier, as there isno ambiguity on what truth assignment to chose.
Here is the inductive step.
Lemma 0.4.5. If Σ is a consistent theory and ϕ is a wff, then Σ ∪ {ϕ} orΣ ∪ {¬ϕ} is consistent.
Proof . Assume that Θ = Σ ∪ {ϕ} is not consistent. By assumption, there is awff ψ such that Θ ` ψ and Θ ` ¬ψ. By ¬i, one has Σ ` ¬ϕ. As Σ is consistent,one cannot have Σ ` ϕ too, so Σ 6` ϕ. So Σ ∪ {¬ϕ} is consistent by Theorem0.3.10.
Lemma 0.4.6. Let Σ be a consistent theory. Then there is a complete consistenttheory Σ containing Σ.
29
Proof . Recall that WFF is countable (Remark 0.1.3, for instance); enumerateWFF = {ϕn : n ∈ N}. We shall extend Σ recursively. Let Σ0 = Σ. Assume Σnhas been constructed. By Lemma 0.4.5, Σn ∪{ϕn} or Σn ∪{¬ϕn} is consistent.If Σn ∪{ϕn} is consistent, let Σn+1 be it; otherwise let Σn+1 = Σn ∪{¬ϕn}. Sothe nth wff ϕn has been taken into account, and Σn+1 is still consistent.
Let Σ = ∪n∈NΣn. Clearly Σ is consistent, and for any wff ϕ, ϕ = ϕn forsome n, so ϕn or its negation is in Σn+1 ⊆ Σ. (It is also clear by consistencythat ϕ and ¬ϕ are not both in Σ.)
0.4.2 Finding a Truth AssignmentA complete consistent theory induces a unique truth assignment, as for anysentence symbol An, either An or ¬An is in Σ. The following lemma checksthat this happens in a consistent way.
Lemma 0.4.7 (truth lemma). Let Σ be a complete consistent theory. Considerthe truth assignment v : A → {F, T} such that v(An) = T iff An ∈ Σ. Then forany wff ϕ, v(ϕ) = T iff ϕ ∈ Σ.
Proof . By induction. The result is clear if ϕ is a sentence symbol! By com-pleteness of Σ, the case ϕ = ¬ψ is immediate. Now suppose that ϕ is ψ1 → ψ2.
• Assume v(ϕ) = T . We show ϕ ∈ Σ.If v(ψ1) = F , then by induction ψ1 6∈ Σ. By completeness of Σ, ¬ψ1 ∈ Σ.Hence Σ ∪ {ψ1} is not consistent; it follows Σ ∪ {ψ1} ` ψ2, and thereforeΣ ` ψ1 → ψ2. By completeness, (ψ1 → ψ2) ∈ Σ.If v(ψ2) = T , then by induction ψ2 ∈ Σ. Therefore Σ ` ψ1 → ψ2; bycompleteness (ψ1 → ψ2) ∈ Σ.
• Assume ϕ ∈ Σ. We show v(ϕ) = T .Otherwise, v(ψ1) = T and v(ψ2) = F . By induction (and completenessof Σ), this shows ψ1 ∈ Σ and ¬ψ2 ∈ Σ. If Σ ` ψ1 → ψ2, then by modusponens, Σ ` ψ2, which contradicts consistency. So Σ 6` ψ1 → ψ2. Thisimplies ϕ 6∈ Σ.
Induction completes the proof. Notice however that proving the theorem withfive connectives would be essentially as hard, though longer.
Proof of the Completeness Theorem (Theorem 0.4.1). Due to Lemma0.4.2, it suffices to show that a consistent theory is satisfiable. Let Σ be aconsistent theory. By Lemma 0.4.6, we can extend Σ to a complete consistentset Σ. We consider the truth assignment v induced by Σ. By Lemma 0.4.7, vsatisfies Σ. In particular v satisfies Σ; Σ is satisfiable.
End of Lecture 6.
Lecture 7 (Compactness of Propositional Logic)
30
0.5 The Compactness Theorem0.5.1 CompactnessCompactness is the general phenomenon of reduction of the infinite to the finite.
Definition 0.5.1 (finite satisfiability). Let Σ be a theory. Σ is finitely satisfiableif for all finite Σ0 ⊆ Σ, Σ0 is satisfiable.
Theorem 0.5.2 (compactness of propositional logic). A theory is satisfiable iffit is finitely satisfiable.
Proof . An implication is trivial: if Σ is satisfiable, then a truth assignmentsatisfying it will also satisfy any finite subset.
Let Σ be a finitely satisfiable theory. Any finite fragment of Σ is satisfiable,hence consistent by soundness (Theorem 0.3.17). So Σ is finitely consistent.But consistency and finite consistency coincide, as consistency is a finite notion(our deductions all have finite length). So Σ is consistent, therefore satisfiableby completeness (Theorem 0.4.1 and Lemma 0.4.2).
Corollary 0.5.3. If Σ |= ϕ, then there is a finite Σ0 ⊆ Σ such that Σ0 |= ϕ.
Proof . By Lemma 0.2.10, Σ∪{¬ϕ} is not satisfiable; by compactness (Theorem0.5.2) there is a finite subset Σ0 ⊆ Σ such that Σ0∪{¬ϕ} is not satisfiable. WithLemma 0.2.10 again, this means Σ0 |= ϕ.
We now give an example of application.
Definition 0.5.4. A graph is k-colorable if there is a function c from the setof vertices to {1, . . . , k} such that if {x, y} is an edge, then c(x) 6= c(y).
Recall that a subgraph of a graph is a subset of the set of vertices equippedwith all possible edges.
Theorem 0.5.5 (Erdös). A countable graph is k-colorable iff all of its finitesubgraphs are k-colorable.
Proof . Let V = {vn : n ∈ N} be an enumeration of the vertices. We addsentence symbols Cv,i for v ∈ V and i ∈ {1, . . . , k} (their meanings will be:c(v) = i, that is v bears the ith color).
Consider the theory made of:
• Cv,1 ∨ · · · ∨ Cv,k for each v ∈ V (meaning: each vertex has a color)
• ∧i6=j 6= Cv,i ∧ Cv,j for each v ∈ V (meaning: no vertex has two colors)
• ∧ki=1 6= Cv,i ∧ Cv′,i for adjacent vertices v, v′ (meaning: no two adjacentvertices bear the same color)
By assumption, the theory is finitely satisfiable. By compactness (Theorem0.5.2), it is satisfiable: there exists a global coloring.
Still not impressed? Wait and see what we shall do with first-order logic. Ex. 0.14, 0.17
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0.5.2 Applications to Decidability*Compactness is very useful is the context of decidability, where one wants todecide whether or not a given wff is a consequence of a theory. We consider thisproblem. For the question to make sense, we need of course to have the data ina form readable by a machine. This is approached by the following definition.
Definition 0.5.6 (decidable). A set is decidable if there is an algorithm whichcan decide whether or not a given object belongs to it.
Example 0.5.7. The set of wff’s is decidable, as there is an algorithm whichdetermines whether a given formula is decidable or not.
Actually if one just wants to list all possible elements of the set, decidabilityis a little too strong, and could be weakened as follow.
Definition 0.5.8 (semi-decidable). A set is semi-decidable if there is an algo-rithmic way to write a list such that sooner or later, every element of the setwill appear on the list.
One also says effectively enumerable for semi-decidable. This does not assertthat there is an algorithm deciding whether or not an object is in the set; all youcan get is an algorithm which will answer “yes” if the object lies in it. Otherwise,the algorithm doesn’t stop, as it scans through an infinite list without findingan answer.
Example 0.5.9. Suppose that you are looking for Jorge Luis Borges in aninfinite phone directory.
• If the phone book is in alphabetical order, you start reading. If you reachByron without passing through Borges, then Jorge Luis Borges is notin the phone book. If you reach Borges, Julio without passing throughBorges, Jorge, then Jorge Luis Borges is not in the phone book. Other-wise you have reached Jorge Luis Borges. The phone book is decidable.
• But now suppose that the phone book is not in alphabetical order. Theonly way to find out if Borges is in the phone book is to read through. Ifyou reach the name Borges, Jorge Luis, then he is in the phone book.Otherwise, you cannot know whether he is not in the directory, or he isbut you haven’t found his name yet. The phone book is semi-decidable.
Remark 0.5.10. A set is decidable iff both it and its complement are semi-decidable.
We now are ready to explain what a “readable” theory is.
Definition 0.5.11 (axiomatization). An axiomatization of a set of wff’s Σ is aset Σ′ such that Σ′ |= Σ and Σ |= Σ′.
Definition 0.5.12 (finitely, decidably, semi-decidably axiomatizable theory).If there exists a finite, decidable, or semi-decidable axiomatization, the theoryis said finitely, decidably, semi-decidably axiomatizable, accordingly.
32
A typical use of Compactness is the following result.
Proposition 0.5.13. If Σ is finitely axiomatizable, then every axiomatizationof Σ contains a finite axiomatization.
Proof . Let Σ0 be a finite axiomatization of Σ; taking the conjunction, we mayassume that Σ0 = ϕ is a single formula.
Let Σ′ be any axiomatization of Σ. Then Σ′ ∪ {¬ϕ} is not consistent, asΣ′ |= Σ |= ϕ. So Σ′ |= ϕ. By compactness, there is a finite subset of Σ′, whichtaking conjunction we may assume to be a formula ψ, such that ψ |= ϕ |= Σ.
We now state the results which interest us.
Corollary 0.5.14. Let Σ be a semi-decidably axiomatisable theory and ϕ a wff.Then there is an algorithm which will answer “yes” if Σ |= ϕ.
Proof . If Σ |= ϕ, then by Compactness (Theorem 0.5.2, or rather Corollary0.5.3), there is a finite Σ0 ⊆ Σ such that Σ0 |= ϕ. We therefore list all finitefragments Σ0 of Σ and apply the finite case algorithm of Theorem 0.2.11 toeach. If Σ |= ϕ, then we shall be answered “yes” in a finite time.
In other words, if Σ admits a semi-decidable axiomatization, then Σ itself,as a set of wff’s, is semi-decidable.
Remark 0.5.15. The algorithm cannot be implemented to say “no”, even ifΣ is decidably axiomatizable. If after a very long time the algorithm hasn’tanswered yes, it might be for two reasons:
• you haven’t waited long enough yet;
• Σ 6|= ϕ (in which case the algorithm will never stop).
Hence, in general, the set of semantic consequences of a semi-decidable Σ isonly semi-decidable (as opposed to the finite case, Theorem 0.2.11, where it isdecidable). However, if the theory is maximal in some sense, there is more.
Corollary 0.5.16. Let Σ be a semi-decidable theory in a decidable language.Assume that for any sentence ϕ, either Σ |= ϕ or Σ |= ¬ϕ. Then there is analgorithm which decides whether Σ |= ϕ or not.
Proof . Run two instances of the algorithm of Corollary 0.5.14: one for ϕ andone for ¬ϕ. One of them will sooner or later answer “yes”.
We do not insist further on decidability; our notion of algorithm is to remaininformal. Decidability and Computability are other branches of MathematicalLogic, with very interesting results.
33
0.5.3 A Topological Proof*Compactness phenomena first arose in the context of topology. Informallyspeaking, a space is compact if whenever one can always find an element meet-ing a finite number of requirements (lying in the intersection of finitely manymembers), then there is an element meeting all requirements simultaneously.And this is exactly what Theorem 0.5.2 says; it actually states the compact-ness of a certain topological space. We therefore give an alternate proof of thecompactness theorem.
Definition 0.5.17 (finite intersection property). A family F has the finiteintersection property if the intersection of any finite number of sets in F isnonempty.
Definition 0.5.18 (compactness). A topological space X is compact if for allfamily of closed sets F with the finite intersection property, ∩F 6= ∅.
For instance, a finite space is always compact.
Theorem 0.5.19 (Tychonoff). A product of compact spaces is compact for theproduct topology.
Corollary 0.5.20. {0, 1}N is compact for the product topology.
We now introduce a topology on the set of truth assignments.
Notation 0.5.21.
• Let S = {F, T}N be the set of truth assignments.
• For every wff ϕ, let Oϕ = {v ∈ S : v(ϕ) = T}.
• Let τ be the topology generated by the Oϕ’s.
Proposition 0.5.22. τ is the product topology on S. S is Hausdorff, compact,and totally disconnected. The Oϕ’s are exactly the clopen sets of S; they forma Boole algebra.
In particular, there is a correspondence between sets of wff’s and closed sets:
Σ 7→ CΣ =⋂ϕ∈Σ
Oϕ
Remark 0.5.23. This correspondence is not injective (as {ϕ} and {¬¬ϕ} definethe same closed set), but it is surjective since any closed set is an intersectionof clopen subsets.
Lemma 0.5.24. Let Σ be a theory and v be a truth assignment. Then v ∈ CΣiff v satisfies Σ.
34
Proof .v ∈ CΣ iff for all ϕ ∈ Σ, v ∈ Oϕ
iff for all ϕ ∈ Σ, v(ϕ) = Tiff v satisfies Σ.
Corollary 0.5.25. Σ is satisfiable iff CΣ 6= ∅.
Proof of the Compactness Theorem, Theorem 0.5.2. Let Σ be a finitelysatisfiable set of wff’s. Consider F = {Oϕ : ϕ ∈ Σ}, a family of closed subsetesof S. As Σ is finitely satisfiable, F has the finite intersection property. Bycompactness of S, ∩F 6= ∅. So CΣ 6= ∅ and Σ is satisfiable.
Remark 0.5.26. The only problem with this proof is that it is not effective,as opposed to the one relying on the completeness theorem. But on the otherhand, it is free of references to a proof theory.
End of Lecture 7.
Lecture 8 (Modal Logic*)
0.6 A Little Modal Logic*We say a word of modal logic. Modal logic is an extension of propositional logictaking necessity into account. (One could also formalize belief, provability, etc.)
Syntactically speaking, this is done by adding adverbs, which will be unaryconnectives. Semantically this is a little more subtle. We need to deal withseveral “worlds”, which will describe the possibilities. In particular a statementwill be necessary if true in all possible worlds.
This notion of possibility requires a treatment of imaginability: some worlds(not necessarily all) are conceivable from others. The notion of a Kripke modelwill be detailed in Definition 0.6.2; for the moment we precise the language.
Definition 0.6.1 (language of modal logic). To the language we add two unaryconnectives: � and �. They denote necessity and possibility, respectively. Weclose WFF under � and �, getting the set WFFmod.
For instance �(�A1 → �A2) is a wff of modal logic, which reads “It isnecessary that the possibility of A1 implies the necessity of A2”. Similarly, ¬�ϕreads “ϕ is contingent”; ¬ � ϕ reads “ϕ is impossible”.
0.6.1 Semantics*The semantics is of course a little more complex than before. As mentioned, weneed a whole collection of worlds, accessible by imagination from each other.
Definition 0.6.2 (Kripke model). A Kripke model is a triple (W,R, v) where
35
• W is a non-empty collection of worlds;
• R is a binary relation on W ; w1Rw2 means that w2 is conceivable in w1;
• v is a function from A×W to {T, F}.
Caution! Truth values depend on the world. v(ϕ) is now meaningless.
Remark 0.6.3. In Definition 0.6.2, we could require specific axioms on theaccessibility relation R in order to formalize a description of what our intuitionsuggests that conceivability means.
Notation 0.6.4. We extend v inductively to WFFmod×W :
• v(An, w) = v(An, w) for each n ∈ N and each world w
• v(¬ϕ,w) = T if v(ϕ,w) = F
• v(ϕ→ ψ,w) = T if v(ϕ,w) = F or v(ψ,w) = T
• v(�ϕ,w) = T if for all world w′ such that wRw′, one has v(ϕ,w′) = T
• v(�ϕ,w) = T if there is a world w′ such that wRw′ and v(ϕ,w′) = T .
The meaning is clear: �ϕ (“ϕ is necessary”) holds in the world w if for anyworld accessible from w, ϕ holds. (An observer from w will then believe that ϕcannot ever fail.) On the other hand, �ϕ (“ϕ is possible”) holds in w if thereis a conceivable world w′ in which ϕ holds: the observer of w will think that ϕcould hold somewhere.
One can of course reduce the language to use the sole modal connective �;clearly � stands for ¬�¬; we implicitely do so and forget about �.
Definition 0.6.5 (satisfaction in modal logic). Σ |= ϕ if for all models (W,R, v)such that Σ is satisfied in all worlds of W , ϕ is also satisfied in all worlds of W .
Example 0.6.6. One has {ϕ} |= �ϕ.
This is somehow problematic, as one should not allow anything similar onthe proof-theoretic side. So we shall have to restrict the completeness theorem(Theorem 0.6.10 below).
The safest is to consider that Example 0.6.6 is not meaningful. We thenagree never to use Σ |= ϕ, but to focus on the case |= ϕ.
0.6.2 Proof Theory*The proof theory is very much alike. We add two rules.
• Kripke’s Rule:Σ ` �(ϕ→ ψ)Σ ` �ϕ→ �ψ
K
36
• Necessitation Rule:` ϕ` �ϕ
N
Caution! There are no assumptions in rule N , i.e. Σ must be the empty set.Otherwise one finds
{ϕ} ` ϕ{ϕ} ` �ϕ
` ϕ→ �ϕ→i
N
in other words, “everything which holds is necessary”, a clear insanity.Recall however that {ϕ} |= �ϕ; in particular, we shall not have “full” com-
pleteness, in the sense that Σ |= ϕ will be stronger than Σ ` ϕ. Completenesswill hold only for Σ = ∅.
Remark 0.6.7. Notice that if we have made specific requirements on the “ac-cessibility” relation R (see Remark 0.6.3), we might need to add a couple ofextra rules.
Example 0.6.8. In Definition 0.6.2, we impose on the accessibility relation tobe reflexive. Then |= (�ϕ)→ ϕ. As there is no way to prove it, we admit it asan axiom, or equivalently we add the following rule:
Σ ` �ϕ
Σ ` ϕ
Theorem 0.6.9 (soundness of modal logic). If Σ ` ϕ, then Σ |= ϕ.
Proof . A quick induction.
• The case of the rules of Propositional Logic is essentially as in Theorem0.3.17. For
Σ ∪ {ϕ} ` ψΣ ` ϕ→ ψ
→i
you might need to introduce W1 = {w in W : v(ϕ,w) = T} and W2 ={w in W : v(ϕ,w) = F}, and prove the claim piecewise.
• Suppose that the last step is` ϕ` �ϕ
N
(Recall that the Necessitation rule has no assumptions).Let (W,R, v) be any Kripke model. We show that for any world w of W ,we have v(�ϕ,w) = T .By induction, |= ϕ, so our Kripke model satisfies ϕ: that is, for any w′ inW , one has v(ϕ,w′) = T . In particular, for any w′ such that wRw′, onehas v(ϕ,w′) = T . Hence v(�ϕ,w) = T .As w is arbitrary, the Kripke model satisfies �ϕ. Since we have taken anyKripke model, we deduce that |= �ϕ.
37
• Now suppose that the last step is
Σ ` �(ϕ→ ψ)Σ ` �ϕ→ �ψ
K
We show Σ |= �ϕ→ �ψ. So let (W,R, v) be a Kripke model satisfying Σ;we show that it satisfies �ϕ→ �ψ.Fiw a world w. We want to show that v(�ϕ → �ψ,w) = T . We maysuppose v(�ϕ,w) = T (as otherwise we are done). By definition, for all w′such that wRw′, one has v(ϕ,w′) = T . We aim at showing v(�ψ,w) = T ,that is v(ϕ,w′) = T for any world w′ accessible from w. Let w′ be suchthat wRw′.By induction, we know Σ ` �(ϕ → ψ). In particular, for any world w′
with wRw′, one has v(ϕ → ψ,w′) = T . As v(ϕ,w′) = T , it followsv(ψ,w′) = T . Since this is true for all w’ accessible from w, one hasv(�ψ,w) = T .One concludes that Σ |= �ϕ→ �ψ.
0.6.3 Completeness*Theorem 0.6.10 (completeness of modal logic). If |= ϕ, then ` ϕ.
Counter-example 0.6.11. This is not true with a non-empty set of assump-tions, as {ϕ} |= �ϕ, but {ϕ} 6` �ϕ.
Proof of Theorem 0.6.10. The proof will rely on the construction of a uni-versal Kripke model (Wun, Run, vun).
• Let Wun be the collection of all maximal consistent modal theories.
• Write wRunw′ iff {ϕ : �ϕ ∈ w} ⊆ w′.
• Let vun : A×Wun → {T, F} be such that vun(An, w) = T iff w ∈ An.
Notice thatWun is not empty by Lemma 0.4.6, or something similar in WFFmod.
Claim (truth lemma; see Lemma 0.4.7). For any ϕ ∈WFFmod and any maxi-mal consistent modal theory w, vun(ϕ,w) = T iff ϕ ∈ w.
Verification: By induction. The only non-trivial case is ϕ = �ψ.
• Suppose ϕ ∈ w. Let w′ be any world such that wRunw′. We assumed �ψ ∈w, so by definition of Run, one has ψ ∈ w′. By induction, vun(ψ,w′) = T .This being true for any w’ with wRunw′, we deduce vun(�ψ,w) = T , thatis vun(ϕ,w) = T .
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• We now suppose vun(ϕ,w) = T ; we show that ϕ ∈ w. Consider Nw ={χ : �χ ∈ w}. We form the modal theory Θ = Nw ∪ {¬ψ}.If Θ is consistent, then it can be extended to a maximal consistent modaltheory w′. Hence Nw ⊆ w′ and by definition of the relation, wRunw′. Byassumption, vun(�ψ,w) = T , so vun(ψ,w′) = T , against ¬ψ ∈ w′.So Θ is not consistent. It follows Nw ` ψ. There are therefore χ1, . . . , χn ∈Nw such that {χ1, . . . , χn} ` ψ. We find
{χ1, . . . , χn} ` ψ` χ1 → (χ2 → (. . . (χn → ψ)...)
→i
` � (χ1 → (χ2 → (. . . (χn → ψ)...))N
` �χ1 → (�χ2 → (. . . (�χn → �ψ)...)K
As �χ1, . . . ,�χn are in w and w is maximal consistent, we see �ψ ∈ w. ♦
We now prove completeness of modal logic. Suppose 6` ϕ. So there is amaximal consistent modal theory w containing ¬ϕ. In particular, vun(¬ϕ,w) =T ; the universal model Wun does not satisfy ϕ. Hence 6|= ϕ.
Remark 0.6.12.
• Of course if we have made restrictions the accessibility relations (and clev-erly added the corresponding deduction rules), we need to prove complete-ness accordingly, that is we must show that the accessibility relation Runon the universal model Wun has the desired properties.
• Notice further that we cannot show that if Σ 6` ϕ, then Σ 6|= ϕ (seeCounter-example 0.6.11). The proof fails for the following reason. Wenaturally restrict our universal model to the setWΣ of maximal consistentextensions of Σ; of course WΣ ⊆Wun, and we take the induced accessibil-ity relation and truth assignment. When trying to prove the truth lemmafor WΣ, we shall form Θ = Σ ∪Nw ∪ {¬ψ}; it is not consistent; we thinkwe are very clever.Hence we find Σ∪{χ1, . . . , χn} ` ψ. But now even if we detach a formulaσ ∈ Σ such that {σ, χ1, . . . , χn} ` ψ, we will end up with �σ → (�χ1 →. . . (�χn → ψ)...), and there is no reason for �σ to be in Σ!
This discussion reveals that the universal modelWΣ satisfies the truth lemmaonly if Σ is closed under �. This yields the following generalization of Theorem0.6.10.
Theorem 0.6.13 (completeness of modal logic revisited). Let Σ be a modaltheory which is closed under �. If Σ |= ϕ, then Σ ` ϕ.
In particular, by a method similar to that of Lemma 0.4.2, we find the dualstatement.
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Theorem 0.6.14 (completeness of modal logic revisited). Let Σ be a modaltheory which is closed under �. If Σ is consistent, then Σ is satisfiable.
Notice that the resulting compactness phenomenon does not require closureunder �.
Corollary 0.6.15 (compactness of modal logic). Let Σ be a modal theory. ThenΣ is satisfiable iff it is finitely satisfiable.
Proof . Suppose Σ is finitely satisfiable; we prove that it is satisfiable (the otherimplication being trivial).
Let Σ′ be the closure of Σ under �: Σ′ is the smallest subset of WFFmodwhich contains Σ and such that ϕ ∈ Σ′ =⇒ �ϕ ∈ Σ′. As for any ϕ ∈WFFmod,{ϕ} ` ϕ, it is clear that Σ′ is finitely satisfiable.
By soundness of modal logic (Theorem 0.6.9), Σ′ is consistent. By complete-ness (Theorem 0.6.14), Σ′ is satisfiable. As Σ ⊆ Σ′, so is Σ.
However interesting epistemologically, one sees on the technical side thatmodal logic is but an easy extension of propositional logic. We shall return tomore mathematical views.
End of Lecture 8.
Lecture 9 (ME1)
End of Lecture 9.
0.7 ExercisesWhen not otherwise specified, we work in classical logic (¬e allowed).
Exercise 0.1. Rewrite the following using only ¬ and →:
1. ϕ ∧ ψ2. ϕ ∨ ψ 3. ϕ↔ ψ
The idea is to show that our connectives are (intuitively speaking) redundant.We shall show that they are formally redundant.
Exercise 0.2. The language has four symbols [, ], {, and }. The collection ofnice formulas is the smallest set of expressions such that:
• the empty/blank expression “” is a nice formula;
• if A,B are nice formulas, [A]{B} is a nice formula.
1. An expression is bracket-balanced if it has as many [’s as ]’s. We definebrace-balanced similarly. Show that every nice formula is bracket-balancedand brace-balanced. (Notice that there are expressions which are bracket-balanced and brace-balanced but not nice.)
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2. An expression is globally balanced if it is both bracket-balanced and brace-balanced, and has as many [’s as {’s. Show that every nice formula isglobally balanced.
3. Show that no proper initial segment of a nice formula is globally balanced.Deduce that no proper initial segment of a nice formula is a nice formula.
Exercise 0.3. Are the following theories satisfiable?
1. Σ1 = {A1, A1 → A2, A2 → A3, A1 ∧A2, A2 ∨ ¬A3}
2. Σ2 = {(A1 → A2)→ A1,¬A1}
3. Σ3 = {A3i ∧A3i+1 → ¬A3i+2 : i ∈ N} ∪ {A2i+1 → ¬A2i+2 : i ∈ N}
Exercise 0.4. Let ϕ be a wff formula which uses as connectives only ∧ and∨. Let v be the truth assignment such that v(An) = T for all n. Show thatv(ϕ) = T .
Exercise 0.5.
1. Show that Σ ` ϕ→ ψ iff Σ ∪ {ϕ} ` ψ.
2. Show that Σ ` ϕ↔ ψ iff Σ ∪ {ϕ} ` ψ and Σ ∪ {ψ} ` ϕ.
Exercise 0.6. Show the following theorems of propositional logic:
1. ` ¬(ϕ ∨ ψ)↔ (¬ϕ ∧ ¬ψ)
2. ` (¬ϕ ∨ ¬ψ)→ ¬(ϕ ∧ ψ)
3. ` ¬(ϕ ∧ ψ)→ (¬ϕ ∨ ¬ψ)
You may use excluded middle only for 3.
Exercise 0.7 (Peirce’s law). Show the following theorem of classical logic:` ((ϕ→ ψ)→ ϕ)→ ϕ.
Exercise 0.8.
1. Write a deduction of ` ϕ→ ¬¬ϕ.
2. Show that ` ¬¬(ϕ ∨ ¬ϕ) (without using ¬e nor excluded middle).
Exercise 0.9 (⊥). Let ⊥ (“perp” or “bottom”) denote contradiction. We in-troduce related deduction rules:
Σ ` ϕ Σ ` ¬ϕΣ `⊥
⊥iΣ `⊥Σ ` ϕ
⊥e
1. Show that Σ is inconsistent iff Σ `⊥.
2. We inductively remove ¬ from wff’s:
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• A′n = An for each sentence symbol;• (¬ϕ)′ = (ϕ′ →⊥);• (ϕ→ ψ)′ = (ϕ′ → ψ′).
We extend the notion of deduction ` to also include the ⊥ rules. Showthat Σ ` ϕ iff Σ′ ` ϕ′.
As we know that other connectives may be coded in terms of ¬ and→, one seesthat all connectives can be coded by → and ⊥. A purely ⊥,→ formulation of¬e is
` ((ϕ→⊥)→⊥)→ ϕ
Exercise 0.10 (Double negation, Excluded middle and Reductio ad Absur-dum). Consider the rules:
Σ ` ¬¬ϕΣ ` ϕ
¬eΣ ∪ {¬ϕ} `⊥
Σ ` ϕ RAA ` ϕ ∨ ¬ϕ EM
Show that all three are equivalent, i.e. taking one as a deduction rule allows toestablish the other two.
Exercise 0.11. Prove the soundness theorem by induction with all five con-nectives.
Exercise 0.12. Let CONS be the property:
Every satisfiable theory is consistent.
Show that soundness is equivalent to CONS.
Exercise 0.13. Let FSAT be the property:
Every finite, consistent theory is satisfiable.
Show that completeness is equivalent to compactness and FSAT.
Exercise 0.14. Let X be a countable set and � a partial ordering on X ((X,�)is a poset):
• ∀x x � x
• ∀x ∀y x � y ∧ y � x→ x = y
• ∀x ∀y ∀z x � y ∧ y � z → x � z.
Show that there is a linear ordering ≤ on X extending �, that is:
• ≤ is an ordering on X
• ≤ is linear: ∀x ∀y x ≤ y ∨ y ≤ x
• ≤ extends �: ∀x ∀y x � y → x ≤ y.
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Hint: When X is finite, an easy induction. When X is countable, use compact-ness of propositional logic.
Exercise 0.15. A map is a countable partition of the plane satisfying certainintuitive properties we do not wish to formalize. It is 4-colourable if there isa function which assigns to each region a colour, such that no two adajacentregions bear the same colour. Show that a countable map is 4-colourable iffeach of its finite submaps is.
Exercise 0.16. A graph is k-colorable if there is a function c from the set ofvertices to {1, . . . , k} such that if {x, y} is an edge, then c(x) 6= c(y). Show thata countable graph is k-colorable iff all of its finite subgraphs are k-colorable.
Exercise 0.17.
1. Let {Sn : n ∈ N} be a collection of finite subsets of N with the followingproperty: for each finite F0 ⊆ N there exists A0 ⊆ N such that for alln ∈ F0, Card(A0 ∩ Sn) = 1. Prove that there exists A ⊆ N such that forall n in N, Card(A ∩ Sn) = 1.
2. Give a counterexample if the Sn’s are not required to be finite.
Exercise 0.18. The purpose of this exercise is to discover intuitionism. Wesay that Σ `IL ϕ if there is a deduction using ¬i, ⊥i, ⊥e, but not ¬e nor itsequivalents RAA and EM.
1. Show `IL ϕ→ (¬¬ϕ). In general the converse does not hold.
2. Show that `IL (¬¬ ⊥)→⊥.
3. Show `IL (¬¬¬ϕ)→ (¬ϕ).
Hence the ¬ operation is not involutive in IL, but quickly becomes involutive.
Exercise 0.19 (Gödel’s translation). We work only with→ and ⊥, as we know(Exercise 0.9) that they are sufficient to encode all connectives.
We define inductively the Gödel translation of a wff
• A∗n = ¬¬An, An a sentence symbol (including ⊥)
• (ϕ→ ψ)∗ = ϕ∗ → ψ∗
1. Show that {ϕ} `CL ϕ∗ and {ϕ∗} `CL ϕ.
2. Show that if `IL ϕ∗, then `CL ϕ.
3. Show that if `CL ϕ, then `IL ϕ∗.
It follows that Intuitionnistic Logic is more subtle than Classical Logic: it hasthe ability to encode all of classical logic, and it perceives differences that CLdoesn’t. (It is weaker only when viewed from CL.)
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Exercise 0.20 (some modal logic). We investigate further the accessibilityrelation. A logic (|=,`) is tautology-complete if whenever |= ϕ, then ` ϕ.
1. Suppose that in any frame, the accessibility relation is reflexive (for w inW , wRw), and that modal logic remains sound and tautology-complete.Show that the following deduction is now valid:
` �ϕ
` ϕ
2. Suppose that the following is now a deduction rule:
` �ϕ
` ϕ
Show that in the canonical model (constructed in the proof of the com-pleteness theorem), the accessibility relation is now reflexive.
3. Deduce that if one considers only relfexive accessibility relations, and al-lows the above rule, then modal logic remains tautology-complete.
4. Same questions with the rule
` �ϕ
` ��ϕ
and the property of transitivity (wRw′ and w′Rw′′ entail wRw′′).
5. Find the proof-theoretic dual of symmetry of (wRw′ entail w′Rw).
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Chapter 1
First-Order Logic (Largo)
First-order Logic, also called Predicate Logic, is more expressive than Propo-sitional Logic: it can mention elements. This requires extending the language;in particular we shall need quantifiers. Up to an abbreviation, we use only one.But this quantifier deeply affects the notion of deduction; we need new rules.On the other hand, the meaning of first-order languages is also richer. We shallintroduce structures, which provide their semantic notion of entailment.
If soundness remains easy to show, completeness of first-order logic will bea major and non-trivial result. Hence, though quite expressive, first-order logicenjoys excellent properties.
In this chapter:
• Define first-order languages (§1.1)
• Introduce the structures associated to first-order languages (§1.2)
• An unpleasant journey to the theory of substitutions (§1.3)
• Extend the notion of deduction to first-order logic (§1.4)
• Prove a soundness theorem for first-order logic (§1.4.3)
• Prove Gödel’s completeness theorem for first-order logic (§1.5)
• Derive compactness (§1.6.2) and find applications (1.6.2)
Lecture 10 (First-Order Languages; Terms and Formulas)
1.1 SyntaxFirst-order is essentially more expressive than Propositional Logic: we even-tually introduce elements. So one will need quantification, but also specific
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elements (constants), properties of elements (relations), and a way to build onelements (functions).
1.1.1 First-order LanguagesWe shall from the start cut down on the number of connectives used; this willmake inductive definitions and proofs way shorter, and is legitimate in view ofthe results of §0.2.3 and 0.3.3.
Definition 1.1.1 (first-order language). A first-order language L consists ofthe following symbols:
• Logical symbols, which are common to all languages:
– “(” and “)”– the connectives ¬ and →– a set V of variables v1, v2, . . . , vn, . . .
– the quantifiers ∀ and ∃– the symbol “=”, which is a binary relation symbol
• Specific symbols, which depend on the language (these form what is calledthe signature of the language):
– a set C of constant symbols– for each n ≥ 1, a set R of n-ary relation symbols– for each n ≥ 1, a set F of n-ary function symbols
Of course one defines the set of expressions of L.
Remark 1.1.2. When giving a first-order language, one does not specify logicalsymbols. In particular, even if it is not mentioned, the symbol “=” is alwayspart of the language.
Remark 1.1.3. It is technically possible to have only relation symbols in thesignature: one replaces constant symbols by unary relation symbols (whichstand for singletons), and n-ary function symbols by n + 1-ary relations (thegraphs of the functions). It is more convenient to allow constant and functionsymbols, but bear in mind that we would be able to work with purely relationallanguages throughout.
Notation 1.1.4. One immediately allows the following abbreviations (E1, E2denote arbitrary expressions):
• E1 ∧ E2 stands for ¬(E1 → ¬E2);
• E1 ∨ E2 stands for (¬E1)→ E2;
• E1 ↔ E2 stands for ¬((E1 → E2)→ ¬(E2 → E1));
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• E1 6= E2 stands for ¬(= (E1, E2)).
This should save us precious amounts of time later, though one could in-troduce five connectives (as we clumsily did for Propositional Logic before theverifications of §0.2.3 and 0.3.3).
Remark 1.1.5. On the other hand, we have two quantifiers. By argumentsressembling those of §0.2.3 and 0.3.3 (see Corollaries 0.2.18 and 0.3.14), we willget rid of ∃, showing that it may be considered as a mere abbreviation. Thiswill be done in §1.4.2 (Corollaries 1.4.8 and 1.4.11).
Example 1.1.6.
• The empty language, or language of pure sets, contains the equality as itsonly relation symbol.
• The language of orderings is Lord = {<}, with < a binary relation symbol(we omit to mention =).
• The language of groups is Lgrps = {1, ·,−1 } where 1 is a constant symbol,· is a binary function symbol, −1 is a unary function symbol.
• The language of rings is Lrings = {0, 1,+,−, ·} where 0 and 1 are constantsymbols, +, −, · are binary function symbols.
1.1.2 TermsWe have defined expressions, but only well-formed formulas are of interest. Yetbefore we can define them, we need to explain which combinations of elementsare meaningful.
Definition 1.1.7 (term). The collection of L-terms is the smallest set suchthat:
• every variable is a term;
• every constant symbol is a term;
• if t1, . . . , tn are terms and f is a n-ary function symbol, then f(t1, . . . , tn)is a term.
Exercise 1.1
Notation 1.1.8 (variables occuring in a term). For an L-term t, we define theset Var(t) of variables occuring in t:
• if t is a constant symbol c, then Var(t) = ∅;
• if t is a variable x, then Var(t) = {x};
• if t is f(t1, . . . , tn), where t1, . . . , tn are terms and f is a n-ary functionsymbol, then Var(t) = Var(t1) ∪ . . . . . .Var(tn).
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It is quite clear that a variable x occurs in a term t iff the symbol x appearsin the expression t.
Example 1.1.9.
• ·(1, ·(v1, 1)) is a term of Lgrps, which is more conveniently written 1·(v1 ·1).Only v1 occurs in it.
• 2 · v21 − 3 · v2 is a term of Lrings, as we have implicitely made the following
abbreviations: 2 stands for +(1, 1), 3 for +(1,+(1, 1)), and v21 for ·(v1, v1).
The variables v1 and v2 occur in the term.
No meaning (and not much interest) so far! We start building on elementsto create well-formed formulas.
1.1.3 FormulasDefinition 1.1.10 (atomic formula). An atomic formula of L is an expressionof the form R(t1, . . . , tn), where R is an n-ary relation symbol of L and t1, . . . , tnare L-terms.
Example 1.1.11.
• v1 = v2 is an atomic formula of any first-order language (as equality isalways part of the language).
• x2 + 1 = 0 is an atomic formula of Lrings.
Definition 1.1.12 (well-formed formula). The collection WFFL of well-formedformulas of L is the smallest set such that:
• every atomic formula is a wff;
• if ϕ and ψ are wff’s, so are ¬ϕ and ϕ→ ψ;
• if ϕ is a wff and x is a variable, then ∀x ϕ and ∃x ϕ are wff’s.
From now on, “formula” will implicitely mean “well-formed formula”.
Example 1.1.13.
• ∀v1∃v2 (v1 · v2 = 1 ∧ v2 · v1 = 1) is (short-hand for) a wff of Lgrps.
• ∀v1 1 = 0 is a wff of Lrings.
• ∀v1 ∀v1 v1 = v1 is a wff of any first-order language.
Theorem 1.1.14 (unique readability). There is only one way to read a wff.
Proof . Proceed exactly like for propositional logic, Theorem 0.1.9. This re-quires a version of the balanced parenthesing lemma, Lemma 0.1.11.
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Notation 1.1.15 (variables occuring in a formula). For an L-wff ϕ, we definethe set Var(ϕ) of variables occuring in ϕ:
• if ϕ is an atomic formula R(t1, . . . , tn), then Var(ϕ) = Var(t1) ∪ · · · ∪Var(tn);
• if ϕ is ¬ψ, then Var(ϕ) = Var(ψ);
• if ϕ is ψ1 → ψ2, then Var(ϕ) = Var(ψ1) ∪Var(ψ2);
• if ϕ is ∀x ψ or ∃x ψ, with x ∈ V, then Var(ϕ) = Var(ψ) ∪ {x}
Again, it is clear that a variable x occurs in a formula ϕ iff the symbol xappears in the expression ϕ.
It is clear from Example 1.1.13 that when working syntactically we will haveto be extremely cautious with variables. This will give rise to several painfultechnicalities, such as the notion of substitutability, to which the entire §1.3 isunfortunately devoted.
Notice however that no one is stupid enough to make such mistakes at thesemantic level. So once we are done with the syntactic study of first-orderlanguage, we shall happily forget the technical details.
Definition 1.1.16 (free or bound variable). Let ϕ be a wff and x a variable.
• if ϕ is atomic, then x occurs free in ϕ if it occurs in ϕ
• if ϕ is ¬ψ, then x occurs free in ϕ if it occurs free in ψ
• if ϕ is ψ1 → ψ2, x occurs free in ϕ if it occurs free in ψ1 or in ψ2
• if ϕ is ∀y ψ or ∃y ψ, x occurs free in ϕ if x occurs free in ψ and x 6= y.
If x occurs in ϕ but is not free, it is said to occur bound in ϕ.
Example 1.1.17.
• In ∀v1 v1 = v2, v1 is bound and v2 is free.
• The same holds of ∀v1 ∀v1 v1 = v2 (which is unfortunately a wff).
• In (∀v1 v2 = v3)→ (v1 = v2), v1, v2 and v3 occur free.
Hence free means: “we’re still missing something to know what it is about”.
Notation 1.1.18. For a formula ϕ, we let FreeVar(ϕ) denote the set of freevariables in ϕ.
Definition 1.1.19 (sentence). A sentence is a wff with no free variable.
Remark 1.1.20. In a sense, in propositional logic all (well-formed) formulasare sentences, as there are no variables.
Definition 1.1.21 (theory). A theory is a set of sentences.
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This definition is the most natural one (we do not want “unexplained vari-ables”), but we shall need to be very careful.
• Some subtleties around the notion of consistency will arise in §1.4.
• We shall define Σ |= ϕ and Σ ` ϕ for sets of formulas, not of sentences, asthis will be useful later.
Example 1.1.22.
• The theory of infinite sets (in Lsets) is
{∃v1 . . . ∃vn ∧i6=j vi 6= vj : n ∈ N}
• The theory of orderings (in Lord) is{∀v1 ¬(v1 < v1), ∀v1∀v2 (v1 < v2 ∨ v1 = v2 ∨ v2 < v1),
∀v1∀v2∀v3 (v1 < v2 ∧ v2 < v3 → v1 < v3)
}• The theory of groups (in Lgrps) is given by the sentences:
– ∀v1∀v2∀v3 v1 · (v2 · v3) = (v1 · v2) · v3
– ∀v1 (v1 · 1 = v1 ∧ 1 · v1 = 1)– ∀v1 (v1 · v−1
1 = 1 ∧ v−11 · v1 = 1)
• The theory of rings is the Lrings-theory given by the expected axioms.
End of Lecture 10.
Lecture 11 (The Semantics of First-Order Logic)
1.2 SemanticsLet us turn our attention to the meaning a first-order formula should convey.This is a matter of interpreting the non-logical symbols (the signature of thelanguage), but also if necessary the free variables. This section is less clumsybut more important than §1.1.
1.2.1 StructuresDefinition 1.2.1 (L-structure). An L-structureM consists of:
• a non-empty base set M (called the base set or underlying set);
• for each constant symbol c of L, a specific element cM of M ;
• for each n-ary relation symbol R of L, a subset RM of Mn;we require that =M must be the equality on M
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• for each n-ary relation function f of L, a function fM from Mn to M .
One also refers to M as the universe of the structure.
We have thus given a meaning to all symbols of L. All? No! We haveforgotten the meaning of the variables, which prevent the terms from having ameaning, which prevent the formulas from having a meaning! So one must alsospecify the value of the variables.
1.2.2 Parameters and InterpretationsDefinition 1.2.2 (assignment of the variables). LetM be an L-structure. Anassignment of the variables, or “assignment” for short, is a function s from V toM .
An assignment is merely a choice of parameters (it specifies what the vari-ables stand for).
Definition 1.2.3 (interpretation of a term with parameters). Let M be anL-structure and s : V → M be an assignment of the variables. We define theinterpretation inM of a term t with parameters s, denoted s(t).
• s(c) = cM for a constant symbol c;
• s(x) already makes sense for a variable x;
• if t is f(t1, . . . , tn), then s(t) = fM(s(t1), . . . , s(tn)).
Lemma 1.2.4 (univocity of interpretation). LetM be an L-structure, t an L-term and s, s′ : V → M be two assignments such that s|Var(t) = s′|Var(t). Thens(t) = s′(t).
Proof . Clear by induction.
1.2.3 Satisfaction and Semantic ConsequenceDefinition 1.2.5 (satisfaction). Let M be an L-structure and s : V → M bean assignment. We define satisfaction inM of a formula ϕ with parameters s,writtenM |= ϕ[s].
• if ϕ is atomic, say R(t1, . . . , tn), thenM |= ϕ[s] if (s(t1), . . . , s(tn)) ∈ RM
(Recall that the symbol = is always interpreted as the equality on M).
• if ϕ is ¬ψ, thenM |= ϕ[s] ifM 6|= ψ[s].
• if ϕ is ψ → χ, thenM |= ϕ[s] ifM 6|= ψ[s] orM |= χ[s].
• if ϕ is ∃x ψ, thenM |= ϕ[s] if there is an assignment s′ such that s ands′ agree on V \ {x} andM |= ψ[s′].
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• if ϕ is ∀x ψ, then M |= ϕ[s] if it is the case that for any assignment s′such that s and s′ agree on V \ {x},M |= ψ[s′].
The idea is clear. The ∃ clause means that there is a value for x such thatψ holds, the other variables being fixed as in s. The ∀ clause means that if onefreezes all variables y 6= x to s(y) but x assumes all possible values m ∈ M , ψholds inM.
Lemma 1.2.6 (univocity of satisfaction). Let M be an L-structure, ϕ an L-formula, and s, s′ : V → M two assignments such that s|FreeVarϕ = s′|FreeVarϕ.ThenM |= ϕ[s] iffM |= ϕ[s′].
In particular, if ϕ is a sentence, thenM |= ϕ[s] does not depend on s.
Proof . A clear induction.
We naturally write M |= Σ[s] if Σ is a set of wff’s such that for all ϕ ∈ Σ,M |= ϕ[s].
If Σ is a set of sentences, thenM |= Σ[s] does not depend on s; in this caseone says thatM is a model of Σ. We do not insist on this terminology now, aswe shall be working mostly with sets of formulas instead of theories.
Definition 1.2.7 (satisfiability). Let Σ be a set of L-wff’s. Σ is satisfiable ifthere is an L-structureM and an assignment s : V →M such thatM |= Σ[s].
Pay attention to the following. We want our notion of consequence to dealnot only with sentences (which have no free variables), but also with formulaswith free variables. For instance we want to say that x = y does imply x+ 1 =y + 1, even though we have no idea of what x and y are. So we do need tooperate on all wff’s.
Definition 1.2.8 (semantic consequence). Let Σ be a set of formulas, ϕ aformula. Σ |= ϕ if for any L-structure M and any assignment s : V → M , ifM |= Σ[s] thenM |= ϕ[s].
Example 1.2.9.
• Let Tgrp be the theory of groups, ϕ the sentence ∀v1 v1 · v1 = 1, and ψthe sentence ∀v1 ∀v2 v1 · v2 = v2 · v1. Then Tgrp ∪ {ϕ} |= ψ.
• For n ∈ N, let ϕn be the sentence ∃v1 . . . ∃vn ∧i 6=j ¬(vi = vj). For eachn ≥ m, one has ϕn |= ϕm.
End of Lecture 11.
Lecture 12 (Substitutions)
1.3 SubstitutionsWe shall soon give deduction rules extending those of propositional logic. Beforethat, we need more notions about the interplay of terms and formulas.
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1.3.1 SubstitutabilityGiven a formula ϕ and a variable x ∈ Varϕ, we may want to replace x by aterm t. This will typically be the case in our proof theory, when we want todeduce ϕ(t) from ∀x ϕ(x). But one must be extremely careful as the followingexamples show.
Example 1.3.1.
• Consider the formula v1 = 1 and the term f(v1). Replacing, one findsf(v1) = 1, and the replacement procedure stops after one iteration.
• Consider the formula ∀v1 v1 = v2. If we replace v1 by a term which is nota variable, the result no longer is a wff! (“∀1 1 = v1”)We shall therefore substitute only free occurences. The problem is thatwe cannot naively susbtitute all free occurences.
• If in ∃v1 v1 = v2 we replace v2 by v1, we find ∃v1 v1 = v1, which clearlyshould mean something else.
• Similarly, if in ∃v1 v1 = v2 we replace v2 by any term involving v1, wealter the meaning.
These examples suggest that we may replace only certain free variables, butnot all. A problem arises when the replacement bounds a variable of t that wasfree in ϕ before.
Definition 1.3.2 (term substitutable for a variable). A term t is substitutablefor a variable x in a formula ϕ if no variable of Var t becomes bound in ϕ whenone replaces every free occurence of x by t.
Example 1.3.3.
• Let ϕ1 be v1 = v2 and t be any term. Then t is substitutable for v1 in ϕ1;it is also substitutable for v2, and substitutable for v3.
• Let ϕ2 be ∀v1 v1 = v1. Then f(v1) is substitutable for v1 in ϕ2, as thereare no free occurences of v1. It is also substitutable for v2, etc.
• Let ϕ3 be ∀v1 v1 = v2. Then any term t is substitutable for v1 in ϕ3. Onthe other hand, t is substitutable for v2 in ϕ3 iff v1 does not occur in t.For instance, if t = f(c, v1), then writing ∀v1 v1 = t would deeply alterthe structure of ϕ3. If t = f(c, v2), or t = f(c, v3), then replacing v2 by tis perfectly licit.
• Let ϕ4 be (∀v1 v1 = v2) → (v2 = v1) and t be a term. Then t is substi-tutable for v1 in ϕ4. On the other hand, t is substitutable for v2 in ϕ4 iffv1 does not occur in t.
Notation 1.3.4. If t is substitutable for x in ϕ, one writes ϕ[t/x] for theexpression where every free occurence of x is replaced by t. (“/” is read “for”).
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We refrain from writing ϕ[t/x] if t is not substitutable for x in ϕ.
Example 1.3.5. Consider Example 1.3.3 again.
• ϕ1[t/v1] is t = v2, whereas ϕ1[t/v3] is of course v1 = v2.
• ϕ2[f(v1)/v1] is ∀v1 v1 = v1. So is ϕ2[f(v1)/v2].
• ϕ3[f(c, v1)/v1] is ∀v1 v1 = v2.ϕ3[f(c, v3)/v2] is ∀v1 v1 = f(c, v3).“ϕ3[f(c, v1)/v2]” is of course not allowed (not substitutable), as the v1from t would become bound.
• ϕ4[f(c, v1)/v1] is (∀v1 v1 = v2)→ (v2 = f(c, v1)).ϕ4[f(c, v3)/v2] is (∀v1 v1 = f(c, v3))→ (f(c, v3) = v1).“ϕ4[f(c, v1)/v2]” is not allowed.
Bear in mind that we replace only free occurences!
1.3.2 A Renaming AlgorithmAs we have said, one must be extraordinarily stupid (or wicked) to constructmisleading formulas; intuitively, we always avoid these dangers. Here is why.
Theorem 1.3.6 (renaming). Let ϕ be a wff, t a term, and x a variable. Thenthere is ϕ which differs from ϕ only in the names of bound variables, and suchthat t is substitutable for x in ϕ, and {ϕ} ` ϕ and {ϕ} ` ϕ.
We do not wish to define the notion ` right now (this will be done in §1.4.1),but it will be clear in due time that the renaming procedure described belowmakes ϕ and ϕ “mutually provable”.
Caution! ϕ depends on ϕ, on x, and on t.
Proof . We construct a suitable ϕ inductively.
• If ϕ is atomic, then there is nothing to do: ϕ can be ϕ. Clearly t issubstitutable for x in ϕ.
• If ϕ is ¬ψ, we apply the algorithm to smaller ψ, and take ϕ to be ¬ψ.Again, t is substitutable for x in ϕ.
• If ϕ is ψ1 → ψ2, we take ϕ to be ψ1 → ψ2. ϕ meets the requirements.
• Suppose that ϕ is ∀y ψ, with y ∈ V.If x = y, that is if ϕ is ∀x ψ, we may take ϕ to be ϕ itself. There are nofree occurences of x in ϕ, so t was trivially substitutable for x in ϕ.If x 6= y and y 6∈ Var(t), then no problem will occur from the quantificationon y. We let ϕ be ∀y ψ; clearly t is substitutable for x in ϕ.
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If x 6= y but y ∈ Var(t), then something must be done. We let z be any(or better: the first!) variable not in Var(ψ) ∪ Var(t) ∪ {x}. Notice thatz is then substitutable for y in ψ (as it doesn’t appear in it); so ψ[z/y]makes sense. Let ϕ be ∀z ψ[z/y]. We claim that t is substitutable for xin ϕ: this is clearly the case, as it is true in ψ, and z does not occur in t.
• We use the same method to deal with ∃y ψ.
Notice that if V has been effectively enumerated, then taking the first vari-able not occuring in Var(ψ) ∪ Var(t) ∪ {x} makes ϕ well-defined, and we havean effective algorithm.
Example 1.3.7. Example 1.3.3 again.
• The easiest is to take ϕ1 for ϕ1. One can substitute any term for anyvariable in ϕ1.
• Similarly, ϕ2 can be ϕ2, as any term is substitutable for any variable inϕ2.
• t is substitutable for v1 in ϕ3, so with respect to v1, ϕ3 could be ϕ3.If v1 6∈ Var(t), then t is substitutable for v2 in ϕ3; ϕ3 can still be ϕ3.Now if v1 ∈ Var(t) and we want to susbtitute t for v1 in ϕ3, somethingmust be done. If for instance t is f(v1), then “∀v3 v3 = v2” will do asϕ3. (In this case, ϕ3[t/v1] is “∀v3 v3 = f(v1)”.) On the other hand if t isg(v1, v2, v3), then the ϕ3 given by our algorithm will be ∀v4 v4 = v2 (andϕ3[t/v1] will now be ∀v4 v4 = g(v1, v2, v3)).Notice that ϕ3 does depend on x and t!
• We want to be able to substitute t for v2 in ϕ4; a problem occurs only ifv1 ∈ Var(t), which we assume. Let z be a variable not occuring in t, norequal to v2. Then “(∀z z = v2) → (v2 = v1)” will do as ϕ4. (Of coursethis ϕ4 only fits a certain desired substitution, as z depends on t.)
If we were not interested in the proof theory of first-order logic, we wouldwork permanently “up to renaming”. Unfortunately we will have to be a littlecareful.
1.3.3 Substitutions and SatisfactionWhat happens if we substitute a term to a variable, and then wonder aboutsatisfaction? Fortunately, everything goes as expected. In what follows, weshall consider an assignement s and a term t which we shall substitute to avariable x. This merely has the effect of changing the value of s at variable x -the new value at x being the interpretation of t.
Lemma 1.3.8. Let t, τ be L-terms, s an assignment, x a variable. Let s agreewith s on V \ {x}, and such that s(x) = s(t). Then s(τ) = s(τ [t/x]).
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Proof . Induction on τ . For clarity, let τ denote τ [t/x]. We want to prove thats(τ) = s(τ).
• This is clear if τ is a constant, or a variable not equal to x.
• If τ is x, then τ is t, and s(τ) = s(x) = s(t) = s(τ).
• If τ is f(τ1, . . . , τn), then s(τ) = s(f(τ1, . . . , τn)) = fM(s(τ1), . . . , s(τn)) =fM(s(τ1), . . . , s(τn)) = s(f(τ1, . . . , τn)) = s(τ).
Proposition 1.3.9. Let ϕ be a formula and t a term substitutable for a variablex in ϕ. Let M be an L-structure and s : V → M an assignment. Let s bethe assignement which agrees with s on V \ {x} but on s(x) = s(t). ThenM |= (ϕ[t/x])[s] iffM |= ϕ[s].
Proof . Induction on ϕ. For clarity we let ϕ denote ϕ[t/x]. Hence we want toproveM |= ϕ[s] iffM |= ϕ[s].
• Suppose that ϕ is atomic, say R(τ1, . . . , τn). Then ϕ is the formulaR(τ1[t/x], . . . , τn[t/x]), which we naturally denote R(τ1, . . . , τn). Now us-ing Lemma 1.3.8,
M |= ϕ[s] iff (s(τ1), . . . , s(τn)) ∈ RMiff (s(τ1), . . . , s(τn)) ∈ RMiff M |= ϕ[s]
• The case where ϕ is obtained via connectives is trivial.
• We suppose ϕ of the form ∀y ψ.First, if y = x, then ϕ is exactly ϕ (as x does not occur free in ϕ). Thenusing Lemma 1.2.6,
M |= ϕ[s] iff M |= ϕ[s]iff M |= ϕ[s]
So the real case is when y 6= x, which we now suppose. By assumption, tis is substitutable for x in ϕ, and this implies y 6∈ Var(t). Furthermore, tis substitutable for x in ψ and ϕ is ∀y ψ.
– Suppose that M |= ϕ[s]; we show that M |= ϕ[s], that is M |=(∀y ψ)[s]. Let s′ be any assignment agreeing with s except on y. Weneed to show thatM |= ψ[s′].By assumption,M |= ϕ[s], which meansM |= (∀y ψ)[s]. Let s′ agreewith s except on y, and such that s′(y) = s′(y). By assumption,M |= ψ[s′].We note that s′ and s′ agree but on x, and s(t) = s(x) = s′(x). Nowy does not appear in t, and s and s′ agree except on y, so by Lemma1.3.8 s′(t) = s(t) = s′(x). As M |= ψ[s′], it follows by induction,M |= ψ[s′]. As s′ is arbitrary on y but coincides with s everywhereelse, this meansM |= (∀y ψ)[s], that isM |= ϕ[s].
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– Now suppose M |= ϕ[s]. We want to show that M |= ϕ[s], that isM |= (∀y ψ)[s]. So let s′ be any assignement agreeing with s excepton y. Let s′ agree with s except on y, and such that s′(y) = s′(y).Notice that s′(x) = s(x) = s(t) = s′(t), as y 6∈ Var(t), and s and s′agree except on y.We know M |= ϕ[s], so M |= (∀y ψ)[s]. In particular, M |= ψ[s′].But s′ and s′ agree everywhere but on x, and s′(x) = s′(t). Soby induction, M |= ψ[s′]. As s′ was arbitrarily obtained from s
by changing its value on y, we find M |= (∀y ψ)[s], and thereforeM |= ϕ[s].
• The case of ∃y ψ is similar.
Counter-example 1.3.10. It is extremely important to have t substitutablefor x in ϕ. Consider the case where ϕ is ∀v2 v1 = v2 and t is v2. Clearly t is notsubstitutable for v1 in ϕ. We let ϕ denote the sentence ∀v2 v2 = v2, resultingfrom an illegitimate replacement.
It is the case that for any L-structureM and assignment s : V →M one willhaveM |= ϕ[s]. But it will not be the case in general thatM |= ϕ[s]. Reasoningas in the proof of Proposition 1.3.9, one does have s′(x) = s(x) = s(t), but nowy occurs in t, and there is no reason why s(t) = s(y) and s′(y) should agree!
End of Lecture 12.
Lecture 13 (Deductions; Simplifying the Language)
1.4 Deductions and Soundness1.4.1 DeductionsWe already know reasonable deduction rules for ¬ and →. In order to extendnatural deduction to first-order logic, we must explain the roles of =, ∀, and ∃.
• Equality requires new axioms, as we want the following intuitively obviousproperties to be deducible.
t = t=
t1 = t2 → (t[t1/x] = t[t2/x])=
t1 = t2 → (ϕ[t1/x]↔ ϕ[t2/x])=
where x is a variable, ϕ is a wff, t, t1, t2 are terms with t1 and t2 sub-stitutable for x in ϕ. (This class of axioms could be made considerablysmaller, but it is not our goal.) Ex. 1.10, 1.11
• ∀ adds deduction rules. One may infer ∀x ϕ from ϕ if x does not appear inth assumption. On the other hand, using a universal quantifier amountsto substituting to a special case.
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• ∃ adds dual deduction rules. One shows an existential by providing anexample, and uses one by introducing a witness.
Exercise 1.9
Of course we shall prove that these rules may be regarded as short-cuts,provided ∃ is considered as an abbreviation for ¬∀¬ (this will be done in§1.4.2).
One could now finish the proof of the renaming algorithm (Theorem 1.3.6)by showing that {ϕ} ` ϕ and {ϕ} ` ϕ. Exercise 1.14
Remark 1.4.1. We must slightly update the definition of consistency. Why?Consider {¬(x = x)}. If no reference to equality is made, then this set offormulas is consistent, but yields inconsistency as soon as one adds the equalityaxioms (which we won’t renounce anyway).
So we redefine consistency as consistency when one adds the axioms of equal-ity (in practice no ambiguity arises).
Proposition 1.4.2.
• ` ∀x ∀y (x = y → y = x)
• ` ∀x ∀y ∀z (x = y ∧ y = z → x = z)
Proof . Let ϕ be the formula z = x, in which we shall replace z by x, or y.Notice that ϕ[x/z] is x = x, but ϕ[y/z] is y = x.
We find the deduction
x = y x = y → ϕ[x/z]→ ϕ[y/z]x = x→ y = x x = x
y = x
Introducing the implication and quantifying, we do get
` ∀x ∀y (x = y → y = x)
Transitivity is an exercise. Exercise 1.13
1.4.2 Simplifying the LanguageWe now explain briefly why two quantifiers are obviously redundant (as far asthe semantics is concerned). This runs parallel to §0.2.3 and 0.3.3.
Notation 1.4.3 (see Notation 0.2.12). Let WFF′ be the set of wff’s that donot use ∃.
Notation 1.4.4 (see Notation 0.2.13). We inductively translate any wff ϕ intoa formula ϕ′ of WFF′:
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• for atomic ϕ, ϕ′ is ϕ;
• (¬ϕ)′ is (¬ϕ′);
• (ϕ→ ψ)′ is (ϕ′ → ψ′);
• (∀x ϕ)′ is(∀x ϕ′)
• (∃x ϕ)′ = ¬(∀x ¬ϕ′)
Of course ϕ and ϕ′ bear the same meaning; we check that viewing ∃ as anabbreviation doesn’t alter our notions of consequence.
Notation 1.4.5 (see Notation 0.2.15). LetM be an L-structure, s : V →M atruth assignment, and ϕ a wff. We defineM |=′ ϕ[s] inductively.
• if ϕ is atomic, say R(t1, . . . , tn), thenM |= ϕ[s] if (s(t1), . . . , s(tn)) ∈ RM
• if ϕ is ¬ψ, thenM |= ϕ[s] ifM 6|= ψ[s].
• if ϕ is ψ1 → ψ2, thenM |= ϕ[s] ifM 6|= ψ1[s] orM |= ψ2[s].
• if ϕ is ∀x ψ, then M |= ϕ[s] if it is the case that for any assignment s′such that s and s′ agree on V \ {x},M |= ψ[s′].
Notice that this is a subdefinition of Definition 1.2.5, but that |=′ is definedonly on formulas of WFF′.
Notation 1.4.6. Let Θ be a consistent subset of WFF′ and ψ ∈ WFF′. Wewrite Θ |=′ ψ if for any L-structure M and assignment s : V → M such thatM |=′ Θ[s], one hasM |=′ ψ[s].
Lemma 1.4.7 (see Lemma 0.2.16). LetM be an L-structure, s an assignment,and ϕ an L-formula. ThenM |= ϕ[s] iffM |=′ ϕ′[s].
Proof . By induction on ϕ, the only non-immediate case being when ϕ is ∃x ψ.But this case is very easy.
Corollary 1.4.8 (see Corollary 0.2.18). Σ |= ϕ iff Σ′ |=′ ϕ′.
Proof . Obvious from Lemma 1.4.7.
The syntactic side is hardly more subtle.
Lemma 1.4.9 (see Lemma 0.3.12). {ϕ} ` ϕ′ and {ϕ′} ` ϕ.
Proof . Induction again. It clearly boils down to treating the case where ϕ is∃x ψ, so that ϕ′ is ¬∀x ¬ψ′.
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• Notice that x is substitutable for x in ψ′. Using induction, one finds
{∃x ψ, ψ,∀x ¬ψ′} ` ψAx,Wk
.... induction
{∃x ψ, ψ,∀x ¬ψ′} ` ψ′{∃x ψ, ψ,∀x ¬ψ′} ` ∀x ¬ψ′
Ax,Wk
{∃x ψ, ψ,∀x ¬ψ′} ` ¬ψ′∀e
{∃x ψ, ψ} ` ¬∀x ¬ψ′¬i︸ ︷︷ ︸
(∗)
so{∃x ψ} ` ∃x ψ (∗)
{ϕ} ` ϕ′∃e
• We need Reductio ad Absurdum (Theorem 0.3.10) for the converse. Weshow that {ϕ′,¬ϕ} is inconsistent.
{¬∃x ψ, ψ′} ` ψ′Ax,Wk
.... induction
{¬∃x ψ, ψ′} ` ψ{¬∃x ψ, ψ′} ` ∃x ψ
∃i {¬∃x ψ, ψ′} ` ¬∃x ψAx,Wk
{¬∃x ψ} ` ¬ψ′¬i︸ ︷︷ ︸
(∗)
Now as x does not appear free in the assumptions,
{¬∃x ψ} ` ¬ψ′
{¬∃x ψ} ` ∀x ¬ψ′∀i
It follows that {¬ϕ,ϕ′} is inconsistent, whence {ϕ′} ` ϕ.
Of course the real problem is to show that one may drop the ∃ rules withoutaffecting the deductive strength.
Notation 1.4.10 (see Notation 0.3.13). Write Σ `′ ϕ if there is a deductionwhich does not use the ∃ rules.
Corollary 1.4.11 (see Corollary 0.3.14). Let Σ be a theory and ϕ a wff. ThenΣ ` ϕ iff Σ′ `′ ϕ′.
Proof . The implication Σ′ `′ ϕ′ ⇒ Σ ` ϕ is very comparable to the proof ofCorollary 0.2.18, thanks to Lemma 1.4.9.
We now show that Σ ` ϕ implies Σ′ `′ ϕ′ by induction on the deduction.The only interesting cases are when the last step is a ∃-rule. There are twopossibilities.
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• Suppose that the last step is:
Σ ` ψ[t/x]Σ ` ∃x ψ
∃i
where t is substitutable for x in ψ. Then.... induction
Σ′ `′ ψ′[t/x]Σ′ ∪ {∀x ¬ψ′} `′ ψ′[t/x]
Wk
{∀x ¬ψ′} `′ ∀x ¬ψ′Ax
Σ′ ∪ {∀x ¬ψ′} `′ ∀x ¬ψ′Wk
Σ′ ∪ {∀x ¬ψ′} `′ ¬ψ′[t/x]∀e
Σ′ `′ ¬∀x ¬ψ′︸ ︷︷ ︸=ϕ′
¬i
so Σ′ `′ ϕ′.
• Now suppose that the last step is:
Σ ` ∃x ψ Σ ∪ {ψ} ` ϕΣ ` ϕ
∃e
where x does not occur free in Σ ∪ {ϕ}. By induction, Σ′ ∪ {ψ′} `′ ϕ′;contraposing (Theorem 0.3.8), Σ′ ∪ {¬ϕ′} `′ ¬ψ′.Now
Σ′ ∪ {¬ϕ′} `′ ¬ψ′
Σ′ ∪ {¬ϕ′} `′ ∀x ¬ψ′∀i
since x does not occur free in Σ′ ∪ {ϕ′}.By induction again, Σ′ `′ ¬∀x ¬ψ′. So Σ′ ∪ {¬ϕ′} is inconsistent, andΣ′ `′ ϕ′ by Reductio ad Absurdum (Theorem 0.3.10).
Hence for us, an existential statement may be proved by showing that auniversal statement does not hold. Intuitively, this is disputable: one has nevergiven an example! For instance, in intuitionistic logic where one does not allow¬e as a deduction rule, ∃ is not equivalent to ¬∀¬. Exercise 1.15
End of Lecture 13.
Lecture 14 (Soundness; Completeness (1/2))
1.4.3 SoundnessTheorem 1.4.12 (soundness of first-order logic). If Σ ` ϕ, then Σ |= ϕ.
Proof . By induction on the deduction. Notice that we are extending propo-sitional logic, so weakening, negation, and implication have been dealt within the proof of the propositional logic version (Theorem 0.3.17). Axioms ofequality are easily dealt with. It remains to consider the case of quantification.By Corollaries 1.4.8 and 1.4.11, only ∀ need be examined: ∃ is now a mereabbreviation.
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• Suppose that the last step of the deduction was:
Σ ` ϕΣ ` ∀x ϕ
∀i
where x does not occur free (Definition 1.1.16) in Σ. Then Σ ` ϕ and byinduction, Σ |= ϕ. We show Σ |= ∀x ϕ.Let M be a structure and s : V → M an assignment. We assume M |=Σ[s] and aim at showingM |= (∀x ϕ)[s].Let s′ be any other assignment agreeing with s on V \{x}. We must showM |= ϕ[s′]. As x does not occur free in Σ, M |= Σ[s′] (Lemma 1.2.6).As Σ |= ϕ, this impliesM |= ϕ[s′]. SoM |= ϕ[s′] regardless of the valueassigned to x by s′. SoM |= (∀x ϕ)[s]. As a conclusion, Σ |= ∀x ϕ.
• Now suppose that the last step was:
Σ ` ∀x ϕΣ ` ϕ[t/x]
∀e
where t is substitutable (Definition 1.3.2) for x in ϕ. We shall show Σ |=ϕ[t/x]. For clarity, let ϕ denote ϕ[t/x].Let M be an L-structure, s : V → M any assignment, and supposeM |= Σ[s]. We have to proveM |= ϕ[s].We know Σ ` ∀x ϕ, and by induction Σ |= ∀x ϕ. AsM |= Σ[s], it followsM |= (∀x ϕ)[s]. So for any s′ which agrees with s on V \ {x}, we haveM |= ϕ[s′]. Let s map x to s(t) (and coincide with s elsewhere); againM |= ϕ[s]. By Proposition 1.3.9, M |= ϕ[s]. Hence for any s such thatM |= Σ[s], we haveM |= ϕ[s]; we are done.
Of course even allowing ∃ as a basic quantifier, the proof wouldn’t be anyharder. Exercise 1.16
1.5 CompletenessSoundness (Theorem 1.4.12) was the easy part. We now show the analog ofTheorem 0.4.1 for first-order logic; informally speaking, it expresses that “if itis true, then there is a deduction of it" (in first-order classical logic for naturaldeduction).
This theorem is due to Kurt Gödel (in his PhD).
Theorem 1.5.1 (completeness of first-order logic). If Σ |= ϕ, then Σ ` ϕ.
We adopt the same strategy as for the case of propositional logic (Theorem0.4.1). It suffices to show the following:
Theorem 1.5.2 (completeness of first-order logic, equivalent version). If Σ |=ϕ, then Σ ` ϕ.
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Theorems 1.5.1 and 1.5.2 are equivalent (see Lemma 0.4.2).
Remark 1.5.3. Throughout we shall be working with sets of formulas insteadof theories (which are sets of sentences: Definition 1.1.21). Even if you havea feeling that it would suffice to show Theorem 1.5.2 for theories, there willbe a point in the proof (the fairly technical Lemma 1.5.28 below) in which allformulas must be taken into account.
Such a subtlety was impossible to imagine in propositional logic, where in asense all formulas are sentences!
Remark 1.5.4. The consistent set we start with may be assumed to containthe axioms of equality. This remark is essential; see Remark 1.4.1.
1.5.0 Strategy and WitnessesStarting with a consistent set of first-order wff’s, we shall find an L-structureand an assignment of the variables satisfying it. As in the case of propositionallogic, this will first require extending to a maximal set of formulas (Lemma1.5.18 of §1.5.2), and then finding an L-structure (Lemma 1.5.28 of §1.5.3).
It is not unrealistic to think that the collection of terms will provide thebase set. (Actually we shall need to “factor out” an equivalence relation, that iswork with a set of equivalence classes; pay no attention now to this slight detailwhich will be explained in §1.5.3.) We shall then define an ad hoc structure onthe base set, ensuring that relations and functions behave as required.
For instance, if in the language there are a constant symbol c and a unaryfunction symbol f , we shall have terms c, f(c), f(f(c)), etc., and we know howto interpret c and f . But the issue is that we also have quantifiers; for example,the sentence ∃v1 f(v1) = c, which expresses that f(v1) = c has a solution, mightbe in the theory. The brilliant idea is to add a pointer to such a solution.
First we should simplify the language, following Corollaries 1.4.8 and 1.4.11of §s:firstorderreducinglanguage. We systematically replace ∃ by ¬∀¬.
For each formula ∀x ϕ, if ∀x ϕ fails, we shall “name the flaw”, i.e. introducea new constant cx,ϕ such that
(¬∀x ϕ)→ ¬ϕ[cx,ϕ/x]
Such a formula means: “If ϕ is not always true, it is because of cx,ϕ precisely”.But this will require extending the language with new constants (§1.5.1),
and extending the collection of formulas to force this phenomenon (§1.5.2).This strategy of adding witnesses is due to Leon Henkin.
Remark 1.5.5. We shall only treat the countable case, but some remarks willindicate how to generalize to uncountable languages.
Let Σ be a consistent set of L-formulas containing the axioms of equality.The goal is to find a nice extension of Σ in a language which provides witnesses.
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1.5.1 Expanding the LanguageWe want to introduce witnesses for formulas of the form ¬∀x ¬ϕ. This raisestwo (minor) issues.
• When we add new constants, we expand the language, so we also createnew formulas, and we need even more witnesses!
• When we add new constants, we expand the class of terms, so we alsoexpand the equality axioms and the ∀ rules. Will Σ still be consistent?
We deal with the first issue. The intuitive idea is that for any formula ofthe form ∀x ϕ, we add a new constant symbol cx,ϕ. Of course when we expandthe language, we add new formulas! It thus looks like we need to repeat theconstruction; fortunately the cardinal remains the same.
Lemma 1.5.6. There is a countable extension L′ of L by a set C′ of newconstant symbols cx,ϕ, such that C′ is in bijection with the set of pairs (x, ϕ)where x ∈ V and ϕ is a L′-formula.
Proof . Let L0 = L and WFF0 = WFF(L0). For each pair (x, ϕ) ∈ V ×WFF0,we add a new constant symbol cx,ϕ, getting the language L1.
We repeat the construction with formulas of WFF1 = WFF(L1), and intro-duce new constant symbols, which yields L2. Let L′ = ∪n∈NLn. Clearly L′ hasthe desired property.
Notice that at each step CardLn remains countable, so CardL′ = CardL.
We now deal with the second possible issue - why does Σ remain consistent inL′? The intuitive reason is clear: if from Σ one can deduce a contradiction in L′,then the deduction has used new constant symbols; L says nothing about these,so they behave like free variables. This is expressed by the following theorem.
In the statement below, ϕ[z/c] denotes brute replacement of all occurences ofthe constant c by the variable z (this creates no confusion, as there was certainlyno quantification over the constant c).
Theorem 1.5.7 (generalized generalisation). Let c be a constant symbol occur-ing in ϕ, and suppose Σ ` ϕ where c does not occur in Σ. Then there is avariable z not occuring in ϕ such that Σ ` ∀z ϕ[z/c].
Example 1.5.8. If Σ ` c = 0 and Σ does not mention c, then Σ ` ∀x x = 0.
Proof of Theorem 1.5.7. We write D : Σ ` ϕ to say that D is a deduction(our usual deductive tree). Notice that there is a finite subset Σ0 ⊆ Σ such thatD : Σ0 ` ϕ.
Let z be a variable not occuring in D nor Σ0 (which are finite, but we havecountably many variables!) We replace c by z in each node of D, forming a treeD[z/c]. We show that D[z/c] : Σ0 ` ϕ[z/c].
• If we use an axiom of equality, then substituting z for c remains an axiomof equality!
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• If the step is for instance of the form:
Σ0 ` ψ Σ0 ` ψ → χ
Σ ` χ→e
then by induction Σ0 ` ψ[z/c] and Σ0 ` (ψ → χ)[z/c] which is Σ0 `ψ[z/c]→ χ[z/c]. Hence
Σ0 ` ψ[z/c] Σ0 ` (ψ → χ)[z/c]Σ0 ` χ[z/c]
→e
which completes the step. The ¬i, ¬e, →i rules are as easy.
• Suppose that the last step was of the form:
Σ0 ` ψΣ0 ` ∀x ψ
∀i
where x does not occur free in Σ0. Notice that by choice of z, (∀x ψ)[z/c]is ∀x (ψ[z/c]). It follows
.... induction
Σ0 ` ψ[z/c]Σ0 ` (∀x ψ)[z/c]
∀i
• Suppose that the last step was of the form:
Σ0 ` ∀x ψΣ0 ` ψ[t/x]
∀e
where t is substitutable for x in ψ. By choice of z, (∀x ψ)[z/c] is ∀x (ψ[z/c])and (ψ[t/x])[z/c] is (ψ[z/c])[t/x]. It follows
.... induction
Σ0 ` ∀x ψ[z/c]Σ0 ` (ψ[t/x])[z/c]
∀e
Hence Σ0 ` ϕ[z/c]. As z does not occur free in Σ0, one has Σ0 ` ∀z ϕ[z/c]. SoΣ ` ∀z ϕ[z/c] too.
Remark 1.5.9. Σ0 appears just to make sure that we can provide a new variablez: perhaps Σ did mention all variable names... but we used only finitely manyin the deduction anyway.
Corollary 1.5.10. Σ with the new equality axioms remains consistent in L′.
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Proof . We shall prove a little more: write Σ `′ ϕ if there is a proof usingdeduction rules associated to L′, and Σ ` ϕ if the deduction is carried in L. Weclaim that if Σ, ϕ do not use new constant symbols, then Σ `′ ϕ implies Σ ` ϕ.This will clearly establish L′-consistency of Σ.
So we suppose Σ `′ ϕ, and we assume Σ finite. Working inductively, forevery new constant symbol ci ∈ C′ which appears in the deduction, there isby Theorem 1.5.7 a variable xi such that Σ ` ∀xi ϕ[xi/ci]; in particular Σ `ϕ[xi/ci]. As the ci’s don’t appear in ϕ, we thus have Σ ` ϕ.
Yes! We have found a reasonable language to work in: it has enough wit-nesses, and Σ remains consistent.
Remark 1.5.11. The construction of L′ immediately generalizes to uncount-able L (the construction remains indexed by N; at each stage, CardLn =CardL). Corollary 1.5.10 is still true.
End of Lecture 14.
Lecture 15 (Completeness (2/2))
1.5.2 Extending the TheoryFrom now on we consider Σ as a consistent set of L′-formulas which containsthe L′-axioms of equality.
We shall next prescribe the role of the set of witnesses C′. Let {(xn, ϕn : n ∈N} be an enumeration of the set of pairs (x, ϕ), where x ∈ V is a variable andϕ is an L′-formula.
Notation 1.5.12 (Henkin axioms). Let ηn be the formula:
(¬∀xn ϕn)→ ¬ϕn[cxn,ϕn/xn]
where cxn,ϕnis the first element of C′ which does not occur in {ϕ0, . . . , ϕn} ∪
{η0, . . . , ηn−1}. For simplicity, we write cn instead of cxn,ϕn (there is no risk ofconfusion with the original constants).
ηn means: “if ϕn is not true for all xn’s, it is because of...” and points to aspecific constant symbol, which has never been mentioned before.
Notation 1.5.13. Let Σ′ = Σ ∪ {ηn : n ∈ N}.
We have expanded our consistent set. Is it still consistent?
Lemma 1.5.14. Σ′ is a consistent set of L′-formulas.
Proof . Let Σn = Σ ∪ {ηk : k ≤ n}. We know that Σ0 = Σ is consistent.Suppose that Σn is; we show that Σn+1 is consistent again.
Otherwise, Σn ` ¬ηn+1. Recall that ηn+1 stands for
(¬∀xn+1 ϕn+1)→ ¬ϕn+1[cn+1/xn+1]
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Minimal work shows Σn ` ¬∀xn+1 ϕn+1 and Σn ` ϕn+1[cn+1/xn+1]. Butby construction (Notation 1.5.12), cn+1 does not occur in Σn. So generalizingon the constant (Theorem 1.5.7), we may replace cn+1 by xn+1 and quantify.We find Σn ` ∀xn+1 ϕn+1, which is a contradiction to its consistency.
Σ′ is the basis of our inductive process; it is a consistent set of L′-formulas.We now maximize the set of requirements (see §0.4.1). Unfortunately, as ob-served at the beginning of the proof of completeness (see Remark 1.5.3), wecannot limit ourselves to sentences, but must take all possible formulas intoaccount.
Definition 1.5.15 (V-complete; see Definition 0.4.3). A consistent set of first-order formulas Σ is V-complete if for any formula ϕ, either ϕ or ¬ϕ is in Σ.
Caution! This is way stronger than what complete ought to be. A V-complete(for: variable-complete) set of formulas even prescribes the behaviour of thevariables (v1 = v2 is in it, or v1 6= v2 is in it). The good notion, when workingwith theories, is that of completeness (Definition 1.6.2 below), which is onlyabout sentences.
Remark 1.5.16 (see Remark 0.4.4). If Σ is V-complete, then for any wff ϕ onehas Σ ` ϕ iff ϕ ∈ Σ.
We now make our way to a V-complete set of formulas containing the Henkinaxioms.
Lemma 1.5.17 (see Lemma 0.4.5). If Σ is consistent and ϕ is a wff, thenΣ ∪ {ϕ} or Σ ∪ {¬ϕ} is consistent.
Proof . Very much like Lemma 0.4.5.
Lemma 1.5.18 (see Lemma 0.4.6). Let Σ be a consistent set of wff’s. Thenthere is a V-complete set Σ containing Σ and the Henkin axioms.
Proof . Like Lemma 0.4.6.
Remark 1.5.19. If L′ is not countable, one has to do a little bit of ordinalinduction. If you know about ordinals, you should see that “taking unions atlimit stages” preserves consistency. The construction is exactly the same.
We have reached a maximal set of L′-formulas which contains all axioms anddescribes the role of witnesses (notice that Σ also describes the relations betweenvariables, as in Definition 1.5.15 we took all formulas, not only sentences). It istime to construct a structure.
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1.5.3 Finding a Structure (and Assignment)Recall that from a set Σ, we built a maximal, consistent set of L′-formulasΣ which contains all axioms of L′, for some expansion L′ of L by new con-stant symbols, and the Henkin axioms (Notation 1.5.12), which explain theirbehaviours.
We are very close to satisfying Σ.
Notation 1.5.20. Let T ′ be the set of terms of L′.
This looks like a very clever base set, but here is a minor complication. Ifin the language there are a constant symbol c and a unary function symbol f ,then c, f(c), etc. are terms. But perhaps the formula f(f(c)) = c is in Σ, inwhich case the terms f(f(c)) and c are required to code the same element.
To force this, we identify pairs of terms which are believed to be equal bythe set of conditions.
Notation 1.5.21. Let ∼ be the relation on T ′:
t ∼ t′ if Σ ` t = t′
Lemma 1.5.22. ∼ is an equivalence relation.
Notation 1.5.23. Let M = T ′/ ∼ be the quotient set of T modulo ∼. We usebrackets [.] to denote equivalence classes.
M will be our universe. We have to interpret symbols of L′.
Notation 1.5.24. We turn M into an L′-structure M with the following in-terpretation:
• For c ∈ L′, let cM = [c].
• For an n-ary relation symbol R of L′, let
RM = {([t1], . . . , [tn]) ∈Mn : Σ ` R(t1, . . . , tn)}
• For an n-ary relation symbol R of L′, let
fM([t1], . . . , [tn]) = [f(t1), . . . , f(tn)]
Lemma 1.5.25. This is well-defined.
Proof . There is no ambiguity for the constants. We do the case of a relationsymbol (a function symbol is handled similarly).
Suppose t1 ∼ t′1, . . . , tn ∼ t′n and Σ ` R(t1, . . . , tn); we show that Σ `R(t′1, . . . , t′n). By definition, the formulas “t1 = t′1”, . . . , “tn = t′n” are in Σ. Asit contains the axioms of equality, we have Σ ` R(t1, . . . , tn) → R(t′1, . . . , t′n).By assumption, Σ ` R(t1, . . . , tn). Detaching, we find Σ ` R(t′1, . . . , t′n).
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We have an L-structure, and now need an assignment of the variables; butas they are terms, they already have been taken care of.
Notation 1.5.26. Let s : V →M map x to [x].
Lemma 1.5.27. For any L′-term t ∈ T ′, s(t) = [t].
Proof . Clear for a constant and a variable; clear by induction.
Lemma 1.5.28 (truth lemma; see Lemma 0.4.7). For any L′-formula ϕ, onehasM |= ϕ[s] iff ϕ ∈ Σ.
Proof . Induction on ϕ. Bear in mind that by V-completeness of Σ, for anyformula, ϕ ∈ Σ iff Σ ` ϕ!
• Suppose ϕ is t1 = t2. Then
M |= ϕ[s] iff s(t1) = s(t2)iff [t1] = [t2]iff Σ ` t1 = t2
• Suppose ϕ is R(t1, . . . , tn). Then
M |= ϕ[s] iff (s(t1), . . . , s(tn)) ∈ RMiff ([t1], . . . , [tn]) ∈ RMiff Σ ` R(t1, . . . , tn)
• Suppose ϕ is ¬ψ. Then
M |= ϕ[s] iff M 6|= ψ[s]iff ψ 6∈ Σiff ¬ψ ∈ Σ
• Suppose ϕ is ψ → χ. Then
M |= ϕ[s] iff M 6|= ψ[s] or M |= χ[s]iff ψ 6∈ Σ or χ ∈ Σiff (ψ → χ) ∈ Σ
(We used consistency and V-completeness of Σ.)
• Suppose ϕ is ∀x ψ. The witnesses play their role here. (x, ψ) is some pair(xn, ϕn), and therefore ϕ is the formula ∀xn ϕn. Recall (Notation 1.5.12)that cn denotes the associated witness.
– Assume M |= ϕ[s]; we show that ϕ ∈ Σ. By assumption, M |=(∀xn ϕn)[s]; let s′ agree with s except on xn and such that s′(xn) =cn. By definition of satisfaction,M |= ϕn[s′]. Proposition 1.3.9 thensaysM |= (ϕn[cn/xn])[s]. By induction, ϕn[cn/xn] ∈ Σ. As ηn ∈ Σand Σ is consistent, one has Σ ` ∀xn ϕn. So ϕ ∈ Σ.
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– Assume M 6|= ϕ[s]; we want to show that ϕ 6∈ Σ. By assumption,there is an assignment s′ which agrees with s except on xn and suchthatM |= (¬ϕn)[s′].By construction of M , s′(xn) is some class [t] for a term t. We wouldlike to use Proposition 1.3.9 and say that “M |= (¬ϕn[t/xn])[s]”; theissue is that perhaps t is not sustitutable for xn in ϕn, so the abovedoes not make sense. We need to rename a bit.Let ϕn be a formula in which t is substitutable for xn and such that{ϕn} ` ϕn and {ϕn} ` ϕn (Theorem 1.3.6). Then M |= (¬ϕn)[s′],s′(xn) = [t] = s(t) (Lemma 1.5.27), and t is substitutable for xnin ϕn. The use of Proposition 1.3.9 is now legitimate, and we findM |= (¬ϕn[t/xn])[s].By induction, ϕn[t/xn] 6∈ Σ. By consistency (and as t is substituablefor xn in ϕn), ∀xn ϕn 6∈ Σ either. We are done with substitutions;as ϕn and ϕn prove each other, we find ∀xn ϕn 6∈ Σ. So ϕ 6∈ Σ.
The reader should have another look at remark 1.5.3; it is now clear that wehad to work with all formulas, and not only sentences.
Proof of Gödel’s Completeness Theorem for First-Order Logic. Fix afirst-order language L and a consistent set of L-formulas Σ. We find an L-structure and an assignment satisfying Σ.
We first expand L to a language L′ having many witnesses (Lemma 1.5.6).We then extend Σ to a V-complete set of L′-formulas Σ, which describes thebehavior of the witnesses (Lemma 1.5.18).
We construct an L′-structureM as in Notation 1.5.24 and an assignment sas in Notation 1.5.26. By Lemma 1.5.28,M |= ϕ[s] for all ϕ ∈ Σ′. In particular,M |= Σ[s]: Σ is satisfiable.
End of Lecture 15.
Lecture 16 (Consequences; Compactness; Non-Standard Analysis)
1.6 Consequences and Compactness1.6.1 Decidability*The definitions of §0.5.2 immediately generalize to the case of first-order logic.
Corollary 1.6.1 (see Corollary 0.5.14). Let L be a first-order language (pre-sented in an effective way), Σ a set of wff’s (presented in an effective way) andϕ a wff. Then there is an algorithm which will answer “yes” if Σ |= ϕ.
Proof . The set of axioms of equality is decidable; so we may assume that Σcontains all the axioms. There is an algorithm which produces all theorems of Σ(try all deductions of length n, then all of length n+ 1, etc.). So far we have analgorithm which answers “yes” if Σ ` ϕ. But by soundness and completeness,this is equivalent to Σ |= ϕ.
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We want to generalize Corollary 0.5.16 to first-order logic; this requires defin-ing some maximality condition for a theory. V-completeness (Definition 1.5.15,an ad hoc notion for the proof of the completeness theorem) is obviously toostrong, as one should care only for sentences. The good notion is the following.
Definition 1.6.2 (complete). A first-order theory Σ is complete if for anysentence ϕ, either Σ ` ϕ or Σ ` ¬ϕ.
Notice that Σ is complete iff maximal as a consistent theory. This is moreunderstandable if one closes Σ under consequence, in which case this definitionis the same as the one we gave when proving completeness of propositional logic,Definition 0.4.3.
Caution! Let us insist that as opposed to V-completeness, only sentences aretaken into account. Completeness is much weaker than V-completeness, but isthe only relevant notion when working with theories.
Corollary 1.6.3 (see Corollary 0.5.16). Let Σ be a semi-decidable theory ina decidable language. Assume that Σ is complete. Then there is an algorithmwhich decides whether Σ |= ϕ or not.
Hence, “semi-decidable and complete is decidable”.
1.6.2 CompactnessTheorem 1.6.4 (compactness of first-order logic). Let Σ be a set of first-orderformulas. Then Σ is satisfiable iff Σ is finitely satisfiable.
Proof . If Σ is finitely satisfiable, it is consistent. As it is consistent, it issatisfiable by the equivalent form of completeness.
Caution! Even if you have a clear idea where to satisfy finite fragments of Σ,you may be surprised by a structure satisfying Σ!
Example 1.6.5. Consider an infinite set A, and let the language contain con-stants ca (a ∈ A). The set of formulas Σ = {v1 6= ca : a ∈ A} is clearly finitelysatisfiable in A, as it suffices to take an assignment with s(v1) not one of thefinitely many mentioned a’s. But of course no assignment to A will satisfy allof Σ simultaneously.
Corollary 1.6.6. If Σ |= ϕ, there is a finite subset Σ0 ⊆ Σ such that Σ0 |= ϕ.
Proof . Like Corollary 0.5.3.
One may want to have another look at the notion of finite axiomatizability(Definition 0.5.12).
Example 1.6.7. The theory of an infinite set is not finitely axiomatisable.
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Verification: Suppose it is. Then by Proposition 0.5.13 the natural axiomati-zation {∃x1 . . . ∃xn ∧i 6=j xi 6= xj : n ∈ N} too contains a finite axiomatisation,which reduces to a single sentence {∃x1 . . . ∃xn ∧i 6=j xi 6= xj}. The latter issatisfied by any possibly finite set with at least n elements, a contradiction. ♦
Before moving to an application, let us comment on a historical misunder-standing.
Ce malheureux Théorème de Compacité est entré par la petiteporte, et on dirait que cette modestie originelle lui cause encore dutort dans les manuels de Logique. C’est à mon avis un résultat beau-coup plus essentiel, primordial (et donc aussi moins sophistiqué),que le Théorème de Complétude de Gödel, qui affirme qu’on peut for-maliser la déduction d’une certaine façon en Arithmétique, et c’estune erreur de méthode que de l’en déduire.
The poor Compactness Theorem got in through the back door, andit seems that this initial modesty still mars it in Logic textbooks. It isin my opinion a result way more essential, primordial (and thus also lesssophisticated) than Gödel’s Completeness Theorem, which states that onecan formalize deduction in a certain way in Number Theory, and it is amethodological error to deduce the former from the latter.
Bruno Poizat, Cours de Théorie des Modèles
The truth is that the compactness theorem is a purely semantic property,and has nothing to do with the notion of deduction. It first appeared in dayswhere deductions were a logician’s main preoccupation; this explains its initialmodesty.
There is an alternate proof of compactness in §1.7.
1.6.3 Non-Standard Analysis*The genesis of calculus is a complicated story. However the notion of infinitesi-mals was one of the ways to start a scientific quarrel (and perhaps have a duel)in the xviiith century. In short, are there numbers which are smaller than allpositive reals? (Of course such numbers could not be real numbers themselves.)Can one introduce such convenient fictions, and work with them? Is what weshow about real numbers using infinitesimals true or do we create inconcisten-cies? We provide a xxth century answer to the controversy.
Our goal is to extend R by infinitesimals. First, we agree that R as a fieldinterests us; that is, R as a ring-structure (as it is customary to treat fields asring-structures). We even equip it with the ordering < and the absolute value |.|(these are actually definable in the ring R, but we may put them in the languagefreely). This is the language we are interested in.
But we want a little more. First, we should introduce a new constant foran infinitesimal. Second, we’re not interested in finding an abstract structure;we want one which behaves like R, or even better, which contains R in a nice
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manner. We shall form an ad hoc theory, which will force the structure tobehave.
Consider the language L = {0, 1, <,+,−, ·, |.|} and the ordered field R. Weadd new constants {cr : r ∈ R} (which will stand for elements of R), finding thelanguage LR, and one more new constant α; let L′ = LR ∪ {α} be the resultingfirst-order language. Let Σ be the set containing the following:
• all LR-sentences which hold true in R;
• the formulas {0 < α < cr : r ∈ R>0}(One cannot formally quantify over r in the latter: the cr’s are elementsof the language, but their collection is not expressible in the language!)
The theory Σ thus states that the cr’s have the same relations as the realnumbers they stand for (in particular, some sentences express that we are talkingabout an ordered field); the extra formulas say that α is a positive elementsmaller than any positive real number.
Lemma 1.6.8. Σ is satisfiable.
Proof . A finite fragment of Σ mentions only finitely many cr’s, which we inter-pret in R as the reals they stand for; only finitely many conditions α < cr arementioned, and we interpret α as a positive real number smaller than the cr’sin question. This discussion shows that Σ is finitely satisfiable; actually thatany finite fragment of Σ is satisfiable in R.
By compactness, Σ is satisfiable (but of course, not in R, because R has noinfinitesimals; same phenomenon as in Example 1.6.5).
Notation 1.6.9 (non-standard reals). Let R∗ be an L′-structure satisfying Σ.
R∗ contains an interpretation of α, which is an infinitesimal, and an inter-pretation of the cr’s, which behave like the real numbers they stand for.
Lemma 1.6.10. R injects canonically into R∗.
Proof . Map r to cR∗r (the interpretation of cr in R∗). This is natural enough,and injective alright: if r 6= s ∈ R, then Σ |= cr 6= cs; hence R∗ |= cr 6= cs.
It goes without saying that the mapping is actually a ring homomorphism.We now freely identify R with its image in R∗, and consider that R ⊂ R∗. Theinclusion is of course proper, as αR∗ 6∈ R: it is an infinitesimal!
We now work on retrieving R. We have added infinitesimals, so in a sensethe non-standard line R∗ is thicker; we have also added their inverses, so it isalso longer than R∗.
Notation 1.6.11 (bounded elements and infinitesimals). Consider the subsetsof R∗:
• b = {x ∈ R∗ : there is r ∈ R>0 : |x| < r};
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• o = {x ∈ R∗ : for all r ∈ R>0 : |x| < r}.
b is the set of bounded elements (“not infinitely large”); o is the set ofinfinitesimals (“infinitely close to 0”). Of course o ∩ R = {0}. If you want tohave a better picture of b and o, you may consider that the real line R embedsinto the non-standard line R∗. b is the portion of the non-standard line whichis not incommensurably longer than R. o is the thickness around ◦.
It goes without saying that though nicely defined, these subsets of R∗ arenot (first-order)-definable: the definitions are not expressible in our languageL′.
Lemma 1.6.12. b is a ring, and o is an ideal of b.
Proof . Easy computations.
We shall now define an isomorphism b/o ' R. The construction is fairlyinteresting. Following the same mental picture as before, when one consideronly the commensurable part and factors out the thickness, one retrieves thestandard line R.
Definition 1.6.13 (standard part). For any non-negative x ∈ b, let stx be theleast upper bound of {r ∈ R : r < x}. (We define stx for negative x ∈ b bystx = − st(−x)).
This makes sense, as R is Dedekind-complete. Let x ∈ b be non-negative;then the subset A = {r ∈ R : r < x} of R is non-empty and bounded abovesince x ∈ b. So A has a least upper bound in R; st is well-defined. A little morebasic algebra reveals:
Lemma 1.6.14. st is a ring-homomorphism with kernel o.
Now as st is clearly onto R (it is the identity on R), we deduce that stinduces an isomorphism b/o ' R. In particular, any element x ∈ b can bewritten in a unique way as the sum of a real number stx (its standard part),and an infinitesimal ε ∈ o. (This is of course not true for elements of R∗ \ b,which are infinitely big.)
One can then start doing analysis in a very elegant fashion, whith argumentssuch as: f is continuous at real a ∈ R iff for any infinitesimal ε, f(a+ ε)− f(a)is infinitesimal.
End of Lecture 16.
Lecture 17 (Proof of Compactness by Ultraproducts*)
1.7 An Alternate Proof of Compactness*1.7.1 Filters and Ultrafilters*Definition 1.7.1 (filter). Let X 6= ∅ be a set. F ⊆ P (X) is a filter on X if:
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• ∅ 6∈ F ;
• if A ∈ F and A ⊆ B, then B ∈ F ;
• if A,B ∈ F , then A ∩B ∈ F .
Example 1.7.2. In a topological space, the collection of neighborhoods of agiven point forms a filter.
Remark 1.7.3. Let B be a family of subsets of X having the finite intersectionproperty (Definition 0.5.17). Then there is a filter F such that B ⊆ F .
Verification: Let F = {Y ⊆ X : there exist B1, . . . , Bn in B such that B1 ∩· · · ∩ Bn ⊆ Y }, i.e. F is the smallest family of sets containing every finiteintersection of members of B. This clearly meets the conditions defining a filter.Notice that ∅ 6∈ F by the finite intersection property. ♦
Recall that Y cofinite in X means that X \ Y is finite.
Example 1.7.4. Assume X is infinite. The Fréchet filter is Fre = {Y ⊆ X :Y is cofinite in X}.
Remark 1.7.5. Let F be a filter on X. Consider IF = {X \Y : Y ∈ F}. ThenIF is an ideal of the ring (P (X), ∅, X,M,∩). Dually, every ideal of P (X) givesrise to a filter: there is a bijection between filters on X and ideals of P (X).
The collection of filters on X may be (partially) ordered by inclusion. Thebijection of Remark 1.7.5 is order-preserving.
Definition 1.7.6 (ultrafilter). A maximal filter on X is called an ultrafilter.
Lemma 1.7.7. Let U be a filter on X. Then U is an ultrafilter iff for all A inP (X), A ∈ U or X \A ∈ U .
Proof . Let U be a filter, and bear in mind that U is closed under finite inter-sections.
Suppose that U is an ultrafilter, and let A ∈ P (X). There are three cases.If for all Y ∈ U , A∩Y 6= ∅, then U ∪{A} is a family as in Remark 1.7.3, so it iscontained in a filter. By maximality of U as a filter, it follows A ∈ U . If for allY ∈ U , (X \A)∩Y 6= ∅, then X \A ∈ U similarly. So it remains the case wherethere are Y1, Y2 ∈ U such that A∩Y1 = (X \A)∩Y2 = ∅. But then Y1 ∩Y2 = ∅,a contradiction.
Suppose that for all A ∈ P (X), A ∈ U or X \ A ∈ U . Let U be a filterextending U : we show U = U . Let Y ∈ U . If Y ∈ U we are done. OtherwiseX \ Y ∈ U ⊆ U , so ∅ = Y ∩ (X \ Y ) ∈ U , a contradiction.
Ultrafilters trivially exist.
Definition 1.7.8 (principal ultrafilter). Let a ∈ X. The principal ultrafilteron a is Pa = {Y ⊆ X : a ∈ Y }.
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Remark 1.7.9. If X is finite, then every ultrafilter on X is principal.
In the interesting case (X infinite), do non-principal ultrafilters exist? As-suming AC (or “Zorn’s Lemma”), they do; AC is actually slightly stronger thannecessary to show this. In fact, there is an Ultrafilter Axiom which is inde-pendent of ZF (and weaker than AC) saying that every filter is included in anultrafilter. Applying this to the Fréchet filter on X (Example 1.7.4), we findnon-principal ultrafilters. This is however highly non-constructive.
Lemma 1.7.10 (AC, or “Ultrafilter Axiom”). Every filter is included in anultrafilter.
Remark 1.7.11. F is an ultrafilter iff IF is a maximal ideal. In particular,Lemma 1.7.10 is equivalent to Krull’s Theorem in Ring Theory.
Verification: Clear, as the bijection of Remark 1.7.5 is inclusion-preserving. ♦
Remark 1.7.12. Alternately, an ultrafilter can be viewed as a {0, 1} measureon X (where a set has measure 1 if it is in the ultrafilter, 0 otherwise). Thismeasure is however not σ-additive since for instance if we look at the Fréchetfilter on N each singleton has measure 0 but their union (equal to N) has measure1.
Lemma 1.7.13 (AC). If B has the finite intersection property, there is anultrafilter U such that B ⊆ U .
Proof . As B has the finite intersection property, it is contained in a filter F byRemark 1.7.3. Then Lemma 1.7.10 gives an ultrafilter U extending F .
Remark 1.7.14. There are 22Card X ultrafilters on X. (This theorem, due toHausdorff, is non-trivial.)
1.7.2 Ultraproducts (The Łoś Structure)*Definition 1.7.15 (ultraproduct). Let I 6= ∅ be a set, U be an ultrafilter on I,and for each i ∈ I letMi be an L-structure. We define the ultraproduct of theMi’s (with respect to U), denoted
∏IMi/U . For simplicity, we refer to it as
M∗.The base ofM∗ is the set
∏IMi modulo the equivalence relation
(mi) ∼ (ni) if {i ∈ I : mi = ni} ∈ U
We interpret constants in L by:
cM∗
=[(cMi)i∈I
]We interpret relations by:
RM∗ ([
(m1i )], . . . ,
[(mk
i )])
if{i ∈ I : RMi(m1
i , . . . ,mki )}∈ U
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Similarly for functions:
fM∗ ([
(m1i )], . . . ,
[(mk
i )])
=[(mk+1
i )]
if{i ∈ I : fMi
(m1i , . . . ,m
ki
)= mk+1
i
}∈ U
Checking that these are well-defined is very similar to showing the transitiv-ity of the equivalence relation: if we have two equivalent sequences, then theymust agree on a set in U , which means that the functions or relations must agreeon a set containing a set in U , hence lying in U .
Theorem 1.7.16 (Łoś). Let ϕ be a wff. Let s∗ : V → M∗ be an assignment,say s∗(x) = [(mi,x)i∈I ], and let si(x) = mi,x; each si is an assignment V →Mi.
ThenM∗ |= ϕ[s∗] iff {i ∈ I :Mi |= ϕ[si]} ∈ U .
Proof . By induction on ϕ.
• If ϕ is atomic then it is clear from our definition ofM∗.
• Now suppose that ϕ is ¬ψ. If M∗ |= ¬ψ[s∗], then M∗ 6|= ψ[s∗]. Byinduction, this means that {i ∈ I : Mi |= ψ[si]} 6∈ U , and since U is anultrafilter (so it always contains a set or its complement), this means that{i ∈ I : Mi 6|= ψ[si]} ∈ U , i.e. that {i ∈ I : Mi |= ¬ψ[si]} ∈ U . HenceM∗ |= ϕ[s∗]. All of these steps are reversible, so the iff holds.
• Next assume that ϕ is ψ ∧ χ. Then:
M∗ |= (ψ ∧ χ)[s∗]iff M∗ |= ψ[s∗] andM∗ |= χ[s∗]iff {i ∈ I :Mi |= ψ[si]}, {i ∈ I :Mi |= χ[si]} ∈ Uiff {i ∈ I :Mi |= ψ[si]} ∩ {i ∈ I :Mi |= χ[si]} ∈ Uiff {i ∈ I :Mi |= (ψ ∧ χ)[si]} ∈ U
(We heavily used that U is an ultrafilter here.)
• Finally, assume that ϕ is ∃x ψ.
– IfM∗ |= (∃x ψ)[s∗], there is an assignment (s∗)′ which agrees with s∗except on x, and such thatM∗ |= ψ[(s∗)′]. Say (s∗)′(x) = [(ni)] fora family (ni) ∈ ΠiMi, and for each i ∈ I let s′i agree with si excepton x, where s′i(x) = ni. By induction {i ∈ I : Mi |= ψ[s′i]} ∈ U ,which implies {i ∈ I :Mi |= (∃x ψ)[si]} ∈ U .
– Now suppose that J = {i ∈ I : Mi |= (∃x ψ)[si]} is in U . Fori ∈ J , let ni ∈ Mi and s′i agree with si on V \ {x} where s′i(x) = nibe such that Mi |= ψ[s′i]. For i ∈ I \ J , let s′i = si. We find aresulting assignment (s′)∗ : V → M∗ which differs from s∗ only onx. As J ⊆ {i ∈ I : Mi |= ψ[s′i]} is in U , we find by induction thatM∗ |= ψ[(s′)∗]. HenceM∗ |= ϕ[s∗].
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1.7.3 Alternate Proof of Compactness*Proof of compactness of first-order logic, Theorem 1.6.4. Fix a finitelysatisfiable first-order theory Σ. Let I be the collection of finite subsets of Σ.For each i ∈ I, there is a structureMi satisfying Σi by the definition of finitesatisfiability.
For each ϕ ∈ Σ, let Aϕ = {i ∈ I : Mi |= ϕ}. Then the family {Aϕ :ϕ ∈ Σ} has the finite intersection property, since the intersection of a finitefamily corresponds to Aϕ1∧···∧ϕn
, which must be nonempty since Σ is finitelysatisfiable. By Lemma 1.7.13, there is an ultrafilter U extending {Aϕ : ϕ ∈ Σ}.We let M∗ =
∏Mi/U (i.e. M∗ is the ultraproduct of the Mi’s with respect
to U).Let ϕ ∈ Σ. We have (using Łoś’ Theorem and our definitions):
M∗ |= ϕ ⇔ {i ∈ I :Mi |= ϕ} ∈ U ⇔ Aϕ ∈ U ,
and the latter is true by construction, soM∗ |= Σ.
Remark 1.7.17. This modern and elegant proof of first-order compactnesshas one drawback: it is highly non-constructive, as it relies on the existence ofultrafilters (Lemma 1.7.13). On the other hand, it is purely semantic and doesnot rely on a hidden completeness argument!
Ultraproducts are very important in model theory, if one is not interestedin effectiveness. They appear in important constructions and deep results. Forinstance, one could have realized the non-standard reals as an ultrapower ofR: fix a non-principal ultrafilter U on N, and let R∗ =
∏N R/U . By Łoś’
Theorem, R∗ is an ordered field. If you want to feel an infinitesimal, considerthe equivalence class, modulo U , of the sequence ( 1
n )n∈N: it is positive, andultimately smaller than any fixed positive real number!
End of Lecture 17.
1.8 ExercisesExercise 1.1. State and prove a version of the Unique Readability Theoremfor terms of a first-order language.
Exercise 1.2. Let L be a purely relational first-order language (i.e., the signa-ture consists only of relation symbols). State and prove the Unique ReadabilityTheorem for L-first-order wff’s.
Exercise 1.3. Let ϕ be the wff (∀v1 v1 = v2)→ (∀v2 v2 = v1).
1. Is v3 substitutable for v1 in ϕ? If yes, give ϕ[v3/v1].
2. Is v1 substitutable for v1 in ϕ? If yes, give ϕ[v1/v1].
3. Is v2 substitutable for v1 in ϕ? If yes, give ϕ[v2/v1].
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4. Let t = f(v1, v2). Find a suitable ϕ given by Theorem 1.3.6, when onewishes to replace v1 by t. Then give ϕ[t/v1].
Exercise 1.4. Let L be the empty language (pure equality).
1. Write an axiom saying that there at least n elements.
2. Write an axiom ϕn saying that there exactly n elements.
3. Write a theory T saying that there are infinitely many elements.
4. Let ψ be a sentence. Show that there is a structure satisfying ψ but notT .
Exercise 1.5. Let L = {s}, s a unary function symbol.
1. Write an axiom ϕ saying that s is injective.
2. Write an axiom ψ saying that s is surjective.
3. Classify structures satisfying of {ϕ,ψ} (i.e., describe them all).
Exercise 1.6. Let L be any language, andM be any structure.
1. Assume M finite. Show that there is a sentence ϕ true in M such thatfor any L-structure N , N |= ϕ iff CardM = CardN .
2. Show that this is not true for infiniteM: find two infinite L-structuresMand N which satisfy the same sentences, but with different cardinalities.
Exercise 1.7. Let E be a binary relation symbol and consider the first-orderlanguage L = {E}.
1. Write a (finite!) set of axioms T expressing that E is an equivalencerelation.From now on we’ll implicitely be working with extensions of T (theoriesextending T ). So it makes sense, for us, to speak about equivalence classes(we say classes, for short).
2. Write an axiom saying that every class is reduced to a single element.
3. Write an axiom saying that there is a class with at least n elements.
4. Write an axiom saying that every class has at least n elements.
5. Write an axiom saying that there is exactly one class with at least nelements.
6. Write an axiom saying that there is a class with exactly n elements.
7. Write an axiom saying that there is exactly one class with exactly n ele-ments.
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8. Write an axiom saying that there are exactly m classes with exactly nelements.
9. Write a theory saying that every class is infinite.
10. Is there a theory satisfied exactly by (sets equipped with) an equivalencerelation with an infinite class?
No justification required.
Exercise 1.8. Let 0 be a constant symbol, s a unary function symbol andconsider the following theory:
T = {∀v1 ∀v2 v1 6= v2 → s(v1) 6= s(v2)} ∪ {∀v1 s(v1) 6= 0}∪{∀v1 ∃v2 v1 6= 0→ v1 = s(v2)}∪{∀v1 s ◦ · · · ◦ s︸ ︷︷ ︸
n times(v1) 6= v1 : n ∈ N}
Describe all structures satisfying T .
Exercise 1.9. The language has a unary relation S.
1. Let M be a non-empty set. Prove (in English):
∃x ∈M ∀y ∈M S(x)→ S(y)
2. Write a deduction of:∃x ∀y (S(x)→ S(y))
Exercise 1.10. Suppose the ∀ rules and the axiom v1 = v1.
1. Show that for any variable x, ` x = x.
2. Show that for any L-term t, ` t = t.
Exercise 1.11. Suppose the ∀ rules, the axiom v1 = v1, and for any functionsymbol f the axiom
(v1 = vn+1 ∧ · · · ∧ vn = v2n)→ f(v1, . . . , vn) = f(vn+1, . . . , v2n)
If t is a term and (si) is a family of terms indexed by i ∈ N, we write t[si/vi]for the term obtained from t where for each i, every occurence of vi is replacedby si. (There is also an obvious notion of substitutability.)
Let (si) and (ri) be two families of terms. Proceeding as follows, show` (r1 = s1 ∧ · · · ∧ rn = sn)→ t[ri/vi] = t[si/vi]:
1. Show that if t is a constant, then the claim holds.
2. Show the same if t is a variable.
3. Show the result for any t.
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Exercise 1.12. Suppose the ∀ rules, the axiom v1 = v1, for any function symbolf the axiom
(v1 = vn+1 ∧ . . . vn = v2n)→ f(v1, . . . , vn) = f(vn+1, . . . , v2n)
and for any relation symbol R the axiom
(v1 = vn+1 ∧ · · · ∧ vn = v2n)→ (R(v1, . . . , vn)→ R(vn+1, . . . , v2n)
Sketch a proof (but no deductions, it would be too long!) of the fact thatfor any families of terms (si) and (ri) and any formula ϕ in which they aresubstituable for the vi’s, ` (r1 = s1 ∧ · · · ∧ rn = sn)→ (ϕ[ri/vi]↔ ϕ[si/vi]).
Just decompose the proof into steps, but do not give any details!
Exercise 1.13. Finish the proof of Proposition 1.4.2 by showing that = istransitive, that is show ` ∀x ∀y ∀z (x = y ∧ y = z → x = z).
Exercise 1.14. Let ϕ be a wff, t a term, x a variable, and ϕ given by therenaming algorithm (Theorem 1.3.6) such that t is substitutable for x in ϕ.Show (we haven’t done it yet) that {ϕ} ` ϕ and {ϕ} ` ϕ.
Exercise 1.15. Formalize first-order logic using ∃, and consider ∀ as an abbre-viation. Show that this is consistent syntactically and semantically.
Exercise 1.16. Prove the soundness theorem by induction with both quanti-fiers.
Exercise 1.17 (Gödel translation, continued). We pursue in the vein of Exer-cise 0.19. Define:
• ϕ∗0 = ϕ0 for ϕ0 an atomic formula;
• (∀x ϕ)∗ = ∀x ϕ∗
Prove the same conclusions as in Exercise 0.19.
Exercise 1.18. For n ∈ N, let ϕn be the formula ∃v1 . . . ∃vn ∧i 6=j vi 6= vj . LetΣ = {ϕn}. Show that Σ 6` ϕn+1.
Exercise 1.19. The language is the language of groups.
1. Give a finite theory whose models are all groups.
2. Give a theory whose models are exactly all infinite groups.
3. Find a finite theory whose models are infinite groups (requires some cul-ture).
4. Show that there is no theory whose models are exactly all finite groups.
Exercise 1.20. The language has a binary relation. Prove that there is notheory whose models are all connected graphs.
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Exercise 1.21 (definable sets). LetM be a first-order structure in a first-orderlanguage L.
1. A subset X ⊆M is definable if there is a formula ϕ with FreeVarϕ = {v0}such that for all assignment s: s(v0) ∈ X iffM |= ϕ[s].(Intuitive meaning: m is in X iff ϕ is true of m.)Show that {0} is definable in (N, <). Show that {1} is definable in (N, <).Show that the set of prime numbers is definable in (N, 0, 1,+, ·).
2. A subset X ⊆ M is definable with parameters µ = (µ1, . . . , µk) ∈ Mk
if there is a formula ϕ with FreeVarϕ = {v0, . . . , vk} such that for allassignment of the variables s mapping vi to mi (i = 1 . . . k): s(v0) ∈ X iffM |= ϕ[s].(Intuitive meaning: m is in X iff ϕ(m,µ1, . . . , µk) is true.)Show that {
√π} is definable with parameters π in (R, 0, 1,+,−, ·).
Show that R>0 is definable in (R, 0, 1,+,−, ·).
3. A subset X ⊆ Mn is definable with parameters µ ∈ Mk if there is a for-mula ϕ with FreeVarϕ = {x1, . . . , xn, y1, . . . , yk} such that for all assign-ment of the variables s mapping yi to µi (i = 1 . . . k): (s(v1), . . . , s(vn)) ∈X iffM |= ϕ[s].Show that the function ÷ is definable in (R, 0, 1, ·). (i.e., show that itsgraph is definable.)
4. Bonus questions: R is not definable in (C, 0, 1,+, ·). Z is not definablein (R, 0, 1,+, ·). All definable subsets of C in (C, 0, 1,+, ·) are finite orcofinite.
Exercise 1.22.
1. Show that {k ∈ Z : k is odd } is definable in (Z,+, ·).
2. Show that each element of Q is definable in (R, 0, 1,+,−, ·).
3. Show that {i,−i} is definable in (C, 0, 1,+,−, ·). Show that {i} is not.
4. Show that R is not definable in (C, 0, 1,+,−, ·).
Exercise 1.23 (non-standard models of (N, s)). Let L = {0, s} where 0 is aconstant symbol and s a unary function; let c be another constant symbol andL′ = L ∪ {c}. Let T consist of all sentences true in (N, 0, s) where s is thesuccessor function.
1. Let T ′ = T ∪ {c 6= s ◦ · · · ◦ s︸ ︷︷ ︸n times
: n ∈ N}. Show that T ′ is consistent.
2. Let N′ be a model of T ′. Construct a canonical injective function N ↪→ N′.Is it surjective?
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3. Let X ⊆ N′ be a definable, non-empty subset (definable by a formula withno parameters). Show that X has a least element.
Exercise 1.24. LetM be a finite first-order structure. Let T be the collectionof all L-sentences true inM. Show that all models of T have same cardinal asM. (They are actually all isomorphic toM, whatever that means.) Show thatthis fails ifM is infinite.
Exercise 1.25.
1. Let T be a complete theory. Suppose that T has an infinite model. Showthat all models of T are infinite.
2. Show that this might fail if T is not complete.
3. Give, in a suitable language, an incomplete theory whose models are infi-nite.
Exercise 1.26 (the collection of definable sets). LetM be a first-order struc-ture. Let Dn be the collection of definable subsets of Mn and D the union ofthe Dn’s.
1. Show that each Dn is closed under complement, finite union, finite inter-section.
2. Show that if X,Y ∈ D, then X × Y ∈ D.
3. Let π1 : M2 →M be the projection on the first coordinates. Show that ifX ∈ D2, then π1(X) ∈ D1. Show that if X ∈ D1, then π−1
1 (X) ∈ D2.
Exercise 1.27. Give an example of a language L, an L-structureM and a wffϕ with FreeVarϕ = {v1} such that M |= ∃v1 ϕ but there is no L-term t suchthatM |= ϕ[t/v1].
Exercise 1.28. Suppose that Σ ` ϕ and that P is a predicate symbol whichappears in neither Σ nor ϕ. Show that there exists a deduction of ϕ from Σwhich doesn’t involve P . (Hint: use the completeness theorem.)
Exercise 1.29. Let L be the language of groups, and T be the theory of Z inthis language, that is T = {ϕ an L-sentence such thatZ |= ϕ}.
Let L′ = L ∪ {c}, where c is a new constant symbol.
1. Let n ∈ Z. Write an L′-sentence saying that c is a multiple of n.
2. Let I be a set of prime numbers. Write an L′-theory extending T suchthat only the primes in I divide c.
3. Deduce that there are uncountably many distinct complete L′-theoriesextending T .
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Exercise 1.30 (universal closures). Let ϕ be a formula in a first-order languageL, and let FreeVarϕ = {x1, . . . , xn}. We define the universal closure of ϕ to be
UnClϕ : ∀xn . . . ∀x1 ϕ
1. Give the universal closure of (∀v1 v1 = v2)→ (v1 = v2).
2. We have ordered the free variables by the natural ordering. To con-vince yourself that this is not essential, show that for any formula ψ wthFreeVarψ = {x, y}, one has
{∀x ∀y ψ} ` ∀y ∀x ψ and {∀y ∀x ψ} ` ∀x ∀y ψ
3. Show that {UnClϕ} ` ϕ.
4. Is the converse true?
Exercise 1.31. Let L have the following signature:
• a binary predicate symbol <;
• two constant symbol a and b;
• a unary function symbol f .
Let T be the L-theory with the following axioms:
• ∀v1 ¬(v1 < v1)
• ∀v1 ∀v2 (v1 < v2 ∨ v2 < v1 ∨ v1 = v2)
• ∀v1 ∀v2 ∀v3 (v1 < v2 ∧ v2 < v3)→ (v1 < v3)
• ∀v1 ∀v2 (v1 < v2)→ (∃v3 v1 < v3 ∧ v3 < v2)
• ∀v1 ∃v2 ∃v3 v2 < v1 ∧ v1 < v3
• a < b
• ∀v1 ∀v2 (v1 < v2)→ (f(v1) < f(v2))
Let σ be the sentence : f(a) < b. Prove that T 6` σ and that T 6` ¬σ.
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Chapter 2
Second-Order Logic(Presto)
In this short chapter we briefly describe second-order logic, in which one mayquantify on collections of elements. This requires new symbols ∀2 and ∃2. Thelogic is amzazingly expressive, too expressive to have interesting results.
In this chapter:
• Explain what second-order means.
• Show that it is essentially more expressive than first-order.
• Show that it is essentially too expressive for logic.
Lecture 18 (Second-Order Logic)
Definition 2.0.1 (second-order language). We now expand our existing collec-tion of symbols by the following:
• for each k ≥ 1, a set of k-ary predicate variables Xk1 , X
k2 , . . . , X
kn, . . .
• the quantifiers ∃2 and ∀2
∀2 now quantifies over collections/relations/functions (a relation is a setof n-uples; a function is a certain relation). For instance, X1(a) means thatthe element a is in the predicate X. Or, the formula Y 2(a, b) means that theelements a and b are in relation according to the binary predicate Y . In a word,one may now quantify over predicates.
Caution! The superscript of a variable denotes its arity, that is the Cartesianpower of the base set in which it lives. But the superscript in ∀2 and ∃2 indicatesthe order of the logic.
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One should define terms and wff’s, but we do not wish to repeat thesediscussions, as we shall not try to describe a proof theory. There are two reasons.
• First, we would have to determine suitable axioms for the relationshipbetween predicates and elements, as we determined reasonable axiomsfor equality in the case of first-order logic. This amounts to agreeing onaxioms for membership, that is agreeing on a theory of sets, which is notthe spirit of this course.
• Second, and worse, this appears a little pointless to us, as compactnesswill fail anyhow: hence so would completeness.
We therefore entirely forget about proof theory.
2.1 Compactness failsProposition 2.1.1. There is a second-order formula expressing that a set isinfinite.
Proof . A set is infinite iff it injects into a proper subset. There is a formulasaying that Y 1 is a subset of X1:
subset(Y 1, X1) : ∀x Y 1(x)→ X1(x)
There is a formula saying that Y 1 is a proper subset of X1:
propersubset(Y 1, X1) : subset(X1, Y 1) ∧ (∃x X1(x) ∧ ¬Y 1(x))
There is a formula saying that Γ2 is the graph of a functional relation:
functional(Γ2) : ∀a∀b1∀b2 Γ2(a, b1) ∧ Γ2(a, b2)→ b1 = b2
There is a formula saying that the relation Γ2 is injective:
injective(Γ2) : ∀a1∀a2∀b Γ2(a1, b) ∧ Γ2(a2, b)→ a1 = a2
There is a formula saying that Γ2 defines a function from X1 to Y 1:
function(Γ2, X1, Y 1) : functional(Γ2) ∧(∀x ∃y X1(x)→ (Y 1(y) ∧ Γ2(x, y))
)There is a formula saying that Γ2 defines an injective function from X1 to Y 1:
injection(Γ2, X1, Y 1) : injective(Γ2) ∧ function(Γ2, X1, Y 1)
There is a formula saying that X1 injects into Y :
injects(X1, Y 1) : ∃2Γ2 injection(Γ2, X1, Y 1)
And eventually, the following formula expresses that X1 is infinite:
infinite(X1) : ∃2Y 1 propersubset(Y 1, X1) ∧ injects(X1, Y 1)
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One should compare this to first-order logic, in which “infinite” is not a singleformula, but an infinite theory not finitely axiomatizable (Example 1.6.7). Exercise 2.1
Corollary 2.1.2. No compactness theorem.
Proof . For n ∈ N, let ϕ be the (first-order) sentence: ∃v1 . . . ∃vn ∧i 6=j vi 6= vj ,which expresses that the underlying set has at least n elements. Let ψ be theformula constructed in Proposition 2.1.1 and consider the following second-ordertheory:
T = {ϕ:n ∈ N} ∪ {∀2X1¬ψ(X1)}
Then T is clearly not satisfiable, but is finitely satisfiable!
In particular, regardless of what our proof theory would have been, com-pleteness would have failed.
2.2 Peano ArithmeticWe now deal with a remarkable property of the integers.
Definition 2.2.1 (Peano Arithmetic). Let 0 be a constant symbol and s aunary function symbol. Let PA be the second-order theory in the language{0, s} with the following axioms:
• s is an injective function
• every element not 0 is in the image of s
• there are no cycles, i.e. for each n the axiom:
∀x s ◦ · · · ◦ s︸ ︷︷ ︸n
(x)¬x
• Every subset which contains 0 and is closed under s is the entire base set:
∀2Y 1 (Y 1(0) ∧ (∀y Y 1(y)→ Y 1(s(y)))→(∀x Y 1(x)
)Theorem 2.2.2 (absolute categoricity of PA). All models of Peano Arithmeticare like N (with the natural interpretations of 0 and s as the successor function).
“Like” stands for isomorphic, which we do not wish to define now (this willbe done in model theory, Definition 1’.2.6). It just means that the bijectionpreserves the language; that up to names, the objects are exactly the same.This result will be referred to as absolute categoricity of PA, a terminologywhich will become clearer after the definition of categoricity in model theory(Definition 1’.2.14).
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Proof . LetM |= PA. We find a bijection between N and M which maps 0 to0M and which is compatible with succession. For the sake of clarity, we writen + 1 for sN(n) (succession in N happens to coincide with adding 1, thoughaddition is not part of the language).
Let f(0) = 0M. Inductively, we now define f(n+ 1) = sM(f(n)). This doesdefine a function N → M , which is by definition compatible with s. Clearly fis injective (induction in N; s has no cycles inM).
It remains to show that f is surjective. Consider im f ⊆ M . This subsetcontains 0M and is closed under sM; asM satisfies the axioms of PA, one hasim f = M . So f is surjective; it is a successor-preserving bijection.
Caution! The temptation is strong to wonder whether or not N is a model ofPA. But the integers existed before we started doing logic, and the fact that wehave found a nice description of their behaviour does not change it!
There is no reason to stop. One may quantify over collections of collectionsof elements and introduce ∀3 and so on. Goes without saying that nth-orderlogic remains sound, but does not enjoy compactness.
End of Lecture 18.
Lecture 19 (ME2)
End of Lecture 19.
Exercise 2.1. Find second-order formulas expressing that a set A is:
1. countable
2. uncountable
3. of cardinality ℵ1
4. of cardinality ℵn
5. of cardinality continuum (that is 2ℵ0 , like R).
Hint: for the last one, use the following algebraic fact. There is up to isomor-phism a unique ordered field which is complete.
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Chapter 1’
Some Model Theory(Furioso)
After the failure of compactness in higher logic, we move back to first-orderlogic, which is obviously the only logic worth considering. Model theory is themathematical study of first-order theories; syntactical considerations disappear,if not entirely, at least in spirit. One will still argue by induction on the lengthof formulas, which is a syntactic notion, but there also exists a purely semanticapproach to model-theory, which emphasizes on back-and-forth methods. Forthese reasons, model theory has been called the semantics of predicate calculus.
In this chapter: a first account of model theory
• Elementary equivalence, elementary inclusion (§1’.1)
• Elementary morphisms (§1’.2)
• Löwenheim-Skolem theorems (§1’.3)
• Back-and-forth methods (§1’.4)
• Quantifier elimination (§1’.5.1)
• ω-saturated models and applications (§1’.5.2 and 1’.5.3)
Lecture 20 (Models and Theories; Elementary Equivalence)
1’.0 A Word on Model-TerroristsWe are through with syntax, and will start doing some mathematics. Ourtreatment of terms and formulas has been remarkably accurate as long as we
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have been interested in proof theory; model theory is on the semantic side, so weshall make notations more convenient, and disregard the risk of an ambiguity.
1’.0.1 Formulas and ParametersNotation 1’.0.1 (tuple notation). A finite sequence of objects (a1, . . . , an) ismore conveniently written a (n is then the length of the tuple a).
We shall often abuse notation of Cartesian powers in order to avoid mention-ing lengths. For instance, if a1, . . . , an ∈ M , we shall feel free to write a ∈ Minstead of a ∈Mn.
From now on we assume that every formula we consider has been suitablyconstructed or renamed (for instance by the algorithm of Theorem 1.3.6); thatis, we take for granted that no (∀v1 v1 = v2)→ (v1 = v2) will play us tricks.
Furthermore, we take x, y, z, etc. but also x1, x2, etc. to be variables them-selves (and not names for general variables). In particular x = x1, . . . , xn is atuple of variables.
Notation 1’.0.2 (formula with free variables). We write ϕ(x) to indicate thatthe only free variables of the formula ϕ are among the tuple of variables x.
Hence ϕ(x) means FreeVarϕ ⊆ x; but we may write ϕ(x) even if somevariables of the tuple x are not used. In any case, ϕ (with no x) denotes asentence. We now get rid of assignments of the variables.
Notation 1’.0.3 (formula with parameters). Let ϕ(x) be a formula with freevariables among x, and M an L-structure. Let m = (m1, . . . ,mn) ∈ Mn havesame length as x. We writeM |= ϕ(m) ifM |= ϕ[s] with s : xi 7→ mi.
It is customary to expand the language via constants.
Notation 1’.0.4 (expansion by constants; LA). Let L be a first-order language,and A a set. LA denotes the language obtained by adding to L new constantsymbols ca for elements a of A.
In this case, we shall always interpret ca by a. Under this convention,M |=ϕ(a) iffM |= ϕ(ca1 , . . . , can).
1’.0.2 QuantificationsWe rehabilitate poor ∃ and make it a quantifier again.
Definition 1’.0.5 (quantifier-free formula). A formula is quantifier-free if it hasno quantifiers.
Definition 1’.0.6 (universal formula). A formula is universal if it is of the form∀v1 . . . ∀vn ϕ0, where ϕ0 is quantifier-free.
Definition 1’.0.7 (existential formula). A formula is existential if it is of theform ∃v1 . . . ∃vn ϕ0, where ϕ0 is quantifier-free.
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Be very careful that ∀v1∃v2 . . . is neither universal nor existential.
Definition 1’.0.8 (quantification rank). We define the quantification rank of aformula ϕ.
• if ϕ is atomic, then qrkϕ = 0;
• if ϕ is ¬ψ, then qrkϕ = qrkψ;
• if ϕ is ψ1 → ψ2, then qrkϕ = max(qrkψ1, qrkψ2);
• if ϕ is ∀x ψ, then qrkϕ = qrkψ + 1.
• if ϕ is ∃x ψ, then qrkϕ = qrkψ + 1.
We do not capture the alternation of ∀’s and ∃’s, though a recursion theoristwould feel at some point interested.
1’.0.3 Models and TheoriesWe shall work only with theories (sets of sentences) instead of arbitrary sets offormulas. Indeed, let us introduce a new constant cx for each variable x, and letL′ = L∪{cx : x a variable}. To every (possibly open) L-formula ϕ we associatean L′-sentence ϕ′ by replacing every free occurence of every free variable by thecorresponding constant; notice that subsitutability is not an issue. And a setof L-formulas Σ is satisfiable iff the L′-theory Σ′ = {ϕ′ : ϕ ∈ Σ} is satisfiable.The advantage being that since Σ′ is a set of sentences, it suffices to provide anL′-structure: assignments of the variables are now useless.
As we shall never use truth values again, T denotes a first-order theory.
Definition 1’.0.9 (model). LetM be an L-structure and T be an L-theory. IfM |= T one says thatM is a model of T .
Model-theorists often abuse terminology and say that T is “consistent” for“satisfiable”; by Gödel’s Completeness Theorem (Theorem 1.5.1) this is harm-less anyway. But one should not think that model-theorists are interested insyntax; all they care about is structures. We even abuse further, by sometimesimplicitely defining a theory to be consistent. This leads to questions like: “isthis theory a theory?” (which means: “is this theory consistent?”). No ambi-guity arises in practice.
Also, we freely identify any theory with the set of its consequences (sincethere is no ambiguity on the notion of consequence). This bridges the gapbetween two different definitions of completeness (Definitions 0.4.3 and 1.6.2),which now perfectly agree. Ex. 1’.2, 1’.6
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1’.1 Elementary Equivalence; Inclusion and El-ementary Inclusion
Remark 1’.1.1. “Elementary” is the standard adjective used in model theory;one must be careful with the definitions (elementary equivalence, Definition1’.1.5, and elementary inclusion, Definition 1’.1.13), as the role of parametersis not necessarily obvious. Terminology tends to be confusing; “elementary”captures a way of thought, not a notion.
1’.1.1 Elementary EquivalenceDefinition 1’.1.2 (theory of a structure). LetM be an L-structure. The first-order L-theory ofM, or the theory ofM for short, is ThM = {ϕ :M |= ϕ}.
This amounts to taking only L-sentences which are true inM; if one wantsto allow parameters, we must generalize a little. Recall that LA is the expansionof L by new constant symbols for elements of A (Notation 1’.0.4).
Definition 1’.1.3 (theory of a structure, with parameters). Let M be an L-structure and A ⊆M a set of parameters.
• The theory ofM over A is Th(M, A) = {ϕ(a) ∈ LA :M |= ϕ(a)}.
• The quantifier-free theory of M over A is Th0(M, A) = {ϕ(a) ∈ LA :M |= ϕ(a) and ϕ is quantifier-free}.
In particular, ThM = Th(M, ∅). One may also notice that Th0(M, A) =Th(M, A) ∩ {ϕ ∈ LA : qrk(ϕ) = 0}.
Remark 1’.1.4. If M is an L-structure and A ⊆ M , then Th(M, A) is acomplete LA-theory.
Verification: InM, ϕ(a) is either true or false! ♦
Model theory is interested in the theories of structures. As far as first-orderproperties are concerned, “being the same” amounts to satisfying the samesentences. This is captured by the following, central notion.
Definition 1’.1.5 (elementary equivalence). Let M,N be L-structures. Mand N are elementarily equivalent, denotedM≡ N , if ThM = ThN .
This behaves very much like an equivalence relation (reflexive, symmetric,transitive), except that the class of all L-structures is not, set-theoreticallyspeaking, a set.
Lemma 1’.1.6. A theory T is complete iff for all modelsM, N of T , one hasM≡ N .
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Proof . Recall that we have identified T with the set of its consequences, andthat a theory is complete iff it is maximal as a consistent set of sentences.
Suppose T complete. If M |= T , then T ⊆ ThM. As T is complete, onehas T = ThM; since T = ThN similarly, it followsM≡ N .
Suppose that any two models of T are elementarily equivalent. Let ϕ be asentence such that neither ϕ not ¬ϕ is in T . Then there are modelsM |= T∪{ϕ}and N |= T ∪ {¬ϕ}. In particular, M ≡ N , and this is a contradiction. Soeither ϕ or ¬ϕ is in T .
So the running question: is this theory complete? Has a partial answer: canone show that any two of its models are elementarily equivalent?
1’.1.2 InclusionWe now turn our attention to substructures; the naive notion (Definition 1’.1.7)is not very useful, the interesting form will be elementary inclusion (Definition1’.1.13). As noted in Remark 1’.1.1, terminology might be a bit confusing, as “el-ementary” will not exactly mean the same thing as in “elementary equivalence”(Definition 1’.1.5).
But let us begin with the naive, not-so-useful, notion of inclusion.
Definition 1’.1.7 (substructure). Let M, N be L-structures. Then M is asubstructure of N , writtenM⊆ N , if the following hold:
• M ⊆ N ;
• cM = cN for all constant symbols c ∈ L;
• RM = RN ∩Mn for all n-ary relation symbols R ∈ L;
• fM = fN ∩Mn+1 for all n-ary function symbols R ∈ L.
(fN∩Mn+1 may also be written(fN)|M|Mn , the restriction and corestriction
of fN to M)
In a word,M bears the L-structure induced by restriction of the structureof N to the base M . (In general, if you found two structuresM and N in thesame language, withM ⊆ N but it is not the case thatM⊆ N as L-structures,then presumably there is something wrong.)
Example 1’.1.8.
• A subgroup of a group is a substructure as a group structure.
• A subring of a ring is a subbstructure as a ring structure.
Remark 1’.1.9. Let N be an L-structure and M ⊆ N be a non-empty subset.Then M can be equipped with an L-structureM⊆ N iff the following hold:
• for each constant symbol c, cN ∈M
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• for each function symbol f , M is closed under fN .
The structureM is then uniquely determined.
When studying Löwenheim-Skolem theorems we shall say a little more aboutgenerating substructures (§1’.3.2).
The following is a very healthy exercise.
Remark 1’.1.10. Let I be a non-empty set and (Mi)i∈I be an increasing chainof L-substructures, that is i ≤ j ⇒ Mi ⊆ Mj . Then ∪i∈IMi naturally bearsan L-structureM∗ such that for all i ∈ I,Mi ⊆M∗.
Lemma 1’.1.11 (∃’s go up, ∀’s go down). Let M ⊆ N be an L-structure. Letϕ(m) be a formula with parameters in M .
(i). If ϕ is quantifier-free, thenM |= ϕ(m)⇔ N |= ϕ(m).
(ii). If ϕ is existential, thenM |= ϕ(m)⇒ N |= ϕ(m).
(iii). If ϕ is universal, then N |= ϕ(m)⇒M |= ϕ(m).
Proof .
(i). The case of an atomic formula is by definition. Then a quick induction onthe complexity.
(ii). Induction again. If there are no quantifiers, then this is known. Supposethat ϕ is ∃x ψ(x,m); ψ is existential again. If M |= ∃x ψ(x,m), thenthere is µ ∈ M such thatM |= ψ(µ,m). By induction, N |= ψ(µ,m); soN |= ϕ(m).
(iii). Might be obtained via (ii), or by similar methods. Exercise 1’.3
In general, “existential witnesses need not go down” (dually, “universal prop-erties need not go up”).
Counter-example 1’.1.12.
• Z ⊆ Q as ring-structures, but Q |= ∃x x+ x = 1 though Z doesn’t.
• Q ⊆ R as ring-structures, but R |= ∃x x2 = 1 + 1 though Q doesn’t.
• R ⊆ C as ring-structures, but C |= ∃x x2 = 1 though R doesn’t.
One sees that this notion of inclusion is therefore not very interesting to us.Elementary inclusions (Definition 1’.1.13) will be more relevant. Exercise 1’.7
End of Lecture 20.
Lecture 21 (Elementary Inclusion; Morphisms)
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1’.1.3 Elementary InclusionWe now introduce a better-behaved notion of inclusion (and substructure). Theintuitive idea is that both structures should agree on as many properties aspossible, allowing parameters. Of course in order for the question to makesense, only elements of the small structure can be taken into account.
Definition 1’.1.13 (elementary substructure). Let M ⊆ N be an inclusionof L-structures. M is an elementary substructure of N , written M � N , ifTh(M,M) = Th(N ,M).
Example 1’.1.14.
• (N, <) 6� (Z, <), as N |= ∀x x ≥ 0 but Z |= ∃x x < 0. (This formula uses0 as a parameter.)
• (Q, <) � (R, <) (not necessarily obvious yet).
• Let M ⊆ N be infinite sets, regarded as pure sets. ThenM� N .
• We have observed in Counter-example 1’.1.12 that none of the inclusionsof rings Z ⊆ Q ⊆ R ⊆ C is elementary.
• As rings, Q � C (not necessarily obvious).
• When we constructed non-standard reals in §1.6.3, we considered a the-ory extending Th(R,R), in order to force the inclusion R ⊆ R∗ to beelementary.
In other words, M � N iff M and N have the same LM -theory, iff N |=Th(M,M). This is also equivalent to: for any LM formula ϕ(m) with parame-ters in M ,M |= ϕ(m) iff N |= ϕ(n).
In particularM� N impliesM≡ N , but the converse need not hold.
Counter-example 1’.1.15. Consider N = (Z, <) andM = (2Z, <) ⊆ N . Asthey are clearly isomorphic (Definition 1’.2.6 below), it is clear that they areessentially the same object, and M ≡ N (this will be more convincing later).YetM |= ∀x (x ≤ 0 ∨ x ≥ 2), and N |= ∃x (0 < x < 2). HenceM and N canbe distinguished by the tuple (0, 2).
Lemma 1’.1.16. Let I be a non-empty set and (Mi)i∈I be an increasing el-ementary chain of L-substructures, that is i ≤ j ⇒ Mi � Mj. Let M∗ =∪i∈IMi. Then for all i ∈ I,Mi �M.
Proof . A very healthy exercise. Exercise 1’.1Ex 1’.4, 1’.5
The following gives a criterion for elementarity of inclusions; it should beused mostly in order to go down, that is, when one already has perfectly under-stood the notion of satisfaction in N , and tries to see ifM⊆ N is an elementarysubstructure. (This will typically be the case in §1’.3.2). The statement is moresubtle than it first seems.
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Theorem 1’.1.17 (Tarksi’s test). LetM⊆ N be an inclusion of L-structures.ThenM� N iff for all formula ϕ(m) with parameters in M ,
if N |= ∃x ϕ(x,m) then there is µ ∈M s.t. N |= ϕ(µ,m)
Proof . ⇒ is clear by definition. We show ⇐ by induction. Let ϕ(m) be aformula with parameters m ∈M . We showM |= ϕ(m)⇔ N |= ϕ(m).
• Suppose ϕ atomic. Clearly,M |= ϕ(m) iff N |= ϕ(m).
• Suppose that ϕ is built from shorter formulas using connectives. There isnot much to prove.
• Suppose that ϕ is ∃x ψ.
– Assume that M |= ∃x ψ(x,m). Then there is µ ∈ M such thatM |= ψ(µ,m). By induction, N |= ψ(µ,m), so N |= ϕ(m). Noticethat we did not use the assumption yet.
– Now assume that N |= ∃x ψ(x,m). By assumption, there is µ ∈ Msuch that N |= ψ(µ,m). By induction,M |= ψ(µ,m), and thereforeM |= ϕ(m).
In short, the inclusion is elementary iff all ∃’s go down; yet it is not a goodidea to try to put Tarski’s test in a nutshell, as there are some subtleties.
Remark 1’.1.18.
• Elementary equivalence is a notion involving satisfaction in two structures(M and N ). However, Tarski’s Test reduces to only one notion of sat-isfaction: in N . Notice indeed that the condition is only about truthin N . This will make Tarski’s test especially valuable when constructingelementary substructures (§1’.3.2).
• One should be very careful that in Tarski’s test (Theorem 1’.1.17), one con-siders all formulas of the form ∃x ϕ, where ϕ too might have quantifiers,and not only existential formulas (read Counter-example below).
Counter-example 1’.1.19. Consider (N, <) and (N ∪ {∞}, <), where ∞ de-notes an extra element greater than all natural numbers. Then N ∪ {∞}has a greatest element (this is a first-order sentence), but N doesn’t: clearlyN 6� N ∪ {∞}. But for a purely existential formula ϕ, one has N |= ϕ(n) iffN∪ {∞} |= ϕ(n). (This need not be entirely clear yet, as a proof would rely onback-and-forth methods.)
1’.2 Morphisms, Elementary Morphisms, Cate-goricity
We now have the objects of model-theory; we want to find the suitable notionof morphisms between them.
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1’.2.1 MorphismsNotation 1’.2.1. If m = (m1, . . . ,mn) is a tuple and σ : M → N is a function,we write σ(m) for the tuple (σ(m1), . . . , σ(mn)).
Definition 1’.2.2 (L-morphism). Let M,N be L-structures and σ : M → Nbe a function. σ is an L-morphism if:
• σ(cM) = cN for any constant symbol c;
• m ∈ RM iff σ(m) ∈ RN for any relation symbol R;
• σ(fM(m)) = fN (σ(m)) for any function symbol f .
In short, an L-morphism is a function which is compatible with the interpreta-tions of L inM and N .
In particular, as equality is always among the relations, one sees that an L-morphism is always injective. This is the reason why one also says L-embedding.
Remark 1’.2.3. The composition of two L-morphisms is an L-morphism; IdMis always an L-morphism.
Example 1’.2.4.
• LetM and N be orderings. Then σ : M → N is an {<}-morphism iff itis an order-preserving (injective) function.
• An injective group homomorphism is an Lgrp-morphism.
• Any ring morphism between fields is an Lrng-morphism.
Remark 1’.2.5.
• IfM⊆ N , then the inclusion map M → N is an L-morphism.
• More precisely, let M,N be two structures with M ⊆ N (the under-lying sets). Then M ⊆ N as L-structures iff the inclusion map is anL-morphism.
• LetM,N be L-structures. Then there is an L-embeddingM→N iff Ncan be turned into a model of Th0(M,M) (theory ofM with no quanti-fiers, parameters in M).
One may rush to the definition of an isomorphism.
Definition 1’.2.6 (L-isomorphism).
• A L-morphism is an L-isomorphism if it is surjective.
• M and N are L-isomorphic, written M ' N (L being implicit) If thereis an L-isomorphism between them.
(Recall that an L-morphism is always injective!)
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Remark 1’.2.7. The composition of two L-isomorphisms, the reciprocal map-ping of an L-isomorphism, are L-isomorphisms.
Example 1’.2.8. (Z, <) and (2Z, <) are clearly isomorphic.
Remark 1’.2.5 however suggested that we did not take the good definition inthe first place: morphisms reduce to inclusions, which in general have but littleinterest. So we look for something more reminiscent of the notion of elementaryinclusion.
1’.2.2 Elementary MorphismsDefinition 1’.2.9 (elementary morphism). An L-morphism σ : M → N iselementary if for all formulas ϕ(m) with parameters in M :
M |= ϕ(m) iff N |= ϕ(σ(m))
Stronger than ordinary morphisms which only preserve interpretation, onecould say that elementary morphisms preserve satisfaction.
Remark 1’.2.10. The composition of elementary morphisms is elementary;IdM is always elementary.
The following must be compared with Remark 1’.2.5.
Remark 1’.2.11.
• IfM⊆ N , then the inclusion map M → N is elementary.
• Let M ⊆ N be two L-structures. Then M � N iff the inclusion map iselementary.
• More generally, there is an elementary embedding M → N iff there isM′ � N such that M ' M′ iff N can be turned into a model ofTh(M,M). Exercise 1’.8
As for L-isomorphisms, we had already reached a very nice notion.
Proposition 1’.2.12. An L-isomorphism is elementary.
Proof . Let σ : M → N be a bijective L-morphism. We show that σ is ele-mentary. By induction on a formula ϕ, we show that for any tuple m ∈ M ,M |= ϕ(m) iff N |= ϕ(σ(m)).
If ϕ is atomic, this is clear by definition of a morphism. The case of connec-tives is not hard to deal with.
Suppose ϕ = ∃x ψ. If M |= ϕ(m), then there is µ ∈ M such that M |=ψ(µ,m). By induction, N |= ψ(σ(µ), σ(m)); therefore N |= ϕ(σ(m)). Con-versely, if N |= ϕ(σ(m)), then there is ν ∈ N such that N |= ψ(ν, σ(m)). Sinceσ is surjective, there is µ ∈M such that σ(µ) = ν. Hence N |= ψ(σ(µ), σ(m));by induction,M |= ψ(µ,m), soM |= ϕ(m).
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Corollary 1’.2.13. IfM' N , thenM≡ N .
Proof . Th(M) is the set of properties of the empty tuple ∅ inM. As there isan isomorphism σ, one has that for any sentence,M |= ϕ iff N |= ϕ.
1’.2.3 CategoricityDefinition 1’.2.14 (categoricity). Let T be a first-order theory with infinitemodels and κ an infinite cardinal. T is κ-categorical if it has a model of cardi-nality κ and any two models of T of cardinal κ are isomorphic.
Caution! This should be opposed to “absolute categoricity” of second-orderPeano arithmetic (Theorem 2.2.2), which is about any two models of PA. First-order theories cannot capture cardinalities as second-order logic does (this willbecome clear with the Löwenheim-Skolem theorems, in §1’.3); absolute cate-goricity is impossible for a first-order theory with infinite models. One thereforecompares only models of the same cardinality.
Example 1’.2.15. Here are some categorical theories (we do not prove thesefacts):
• The theory of dense linear orderings is ℵ0-categorical (Theorem 1’.4.4 be-low), but not κ-categorical for any κ ≥ ℵ1 (harder).
• The theory of vector spaces over a finite field is categorical in any cardinal.
• The theory of vector spaces over Q is not ℵ0-categorical, but it is κ-categorical for any κ ≥ ℵ1 (generalize this statement to bigger base fields!).
• The theory of algebraically closed fields of a given characteristic is notℵ0-categorical, but it is κ-categorical for any κ ≥ ℵ1 (hint: transcendencebases).
Categoricity will provide a criterion for completeness of a theory (Corollary1’.3.14). But let us briefly leave the syllabus.
Theorem 1’.2.16 (Morley’s categoricity theorem). A theory categorical in someκ ≥ L is categorical in any κ ≥ L.
Micahel Morley actually proved the countable language case in 1965; SaharonShelah extended it to any language in 1974.
End of Lecture 21.
Lecture 22 (Löwenheim-Skolem Theorems)
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1’.3 Löwenheim-Skolem Theorems1’.3.1 The Substructure Generated by a SetNotation 1’.3.1. If L is finite, we write CardL = ℵ0 anyway.
Notice that there is no risk of confusion if CardL ≥ ℵ1. You may think thatin any case, CardL actually denotes Card WFFL.
Let us start with an L-structure M and a subset A of M . One can easilyfind the smallest substructure ofM containing A: it suffices to close A (unionthe constants) under the functions of the language; the resulting substructurehas cardinality at most CardA+ CardL.
Example 1’.3.2. Let G be a group and S. Then there is a smallest subgroupof G containing S; it is precisely the closure of S under the group multiplicationand inversion. If S is countable, then so is 〈S〉.
Lemma 1’.3.3. Let M be a first-order structure, A ⊆ M . Then there is asmallest L-structure 〈A〉 of M containing A. Moreover, Card〈A〉 ≤ CardA +CardL.
Proof . We first deal with the constants by considering A0 = A ∪ {cM : c ∈L a constant symbol}.
We now write the set of function symbols of the language F =∐n∈N Fn,
where Fn is the set of function symbols of arity exactly n. Let
A1 = A0 ∪⋃n∈N
⋃f∈Fn
fM(An0 )
(The idea is clear. One wishes to add the images of elements of A0 throughfunctions of L. Unfortunately f(A0) does not necessarily make sense, as thearity of the function may vary. The union indexed over N deals with this minor,strictly notational, issue.)
From A1 we build A2 in a similar fashion, and keep going. Eventually theset 〈A〉 = ∪n∈N contains A, all (interpretations of) constants, is closed under(the interpretations of) all functions in the language. We interpret relationsymbols on 〈A〉 by the induced interpretation (i.e., as a subset of M), gettingan L-structure which is a substructure ofM.
Notice that A0 has cardinality at most CardA + CardL; at each stage, weadd at most CardL elements; there are ℵ0 ≤ CardL stages, and this provesCard〈A〉 ≤ CardA+ CardL.
Example 1’.3.4. Consider the case of a subset S of a group G again. ThenS0 = S ∪ {1}, S1 = S0 ∪ S−1
0 ∪ S0 · S0, etc. One does find 〈S〉 = ∪n∈NSn.
Definition 1’.3.5 (substructure generated by a set). LetM be an L-structureand A ⊆M be any set. The substructure generated by A inM is the smallestsubstructure ofM containing A. It coincides with 〈A〉 from Lemma 1’.3.3.
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One can prove that an arbitrary intersection of substructures of M is asubstructure of M again; we have just been generalizing to L a constructionwhich is well-known for groups. It follows that 〈A〉 is the intersection of allsubstructures ofM containing A. Notice in any case that 〈A〉 is canonical, andthat we could construct it explicitely.
All this is very nice, but in general 〈A〉 has no reason to be an elementarysubstructure of M (Definition 1’.1.13), and this is precisely what we would beinterested in. So Lemma 1’.3.3 and Definition 1’.3.5 are not that useful.
1’.3.2 Skolem Functions and the Descending VersionIf we want to find an elementary substructure containing A, then by Tarski’sTest (Theorem 1’.1.17) we ought to add existential witnesses for all formulas.
Recall how existential witnesses played an essential role in the proof of thecompleteness of first-order logic. We now deal with them in a more semanticfashion, adding to the language functions which choose them (this constructionis of course not canonical, and heavily relies on the axiom of choice).
Caution! The substructure we shall construct in Theorem 1’.3.9 will not becanonical.
Definition 1’.3.6 (Skolem functions). Let L be a first-order language. For anyformula ϕ(x, y), where y is an n-uple of variables, we add a new function symbolfϕ, the Skolem function associated to ϕ.
We let LSk = L ∪ {fϕ : ϕ(x, y) an L-formula}.
Definition 1’.3.7 (Skolemization). Let T be an L-theory. The Skolemizationof T is the LSk-theory TSk which is the union of T and the axioms
∀y(∃x ϕ(x, y) → ϕ(fϕ(y), y)
)The meaning is clear: our new axioms say that witnesses are explicitely given
by our new function symbols.
Lemma 1’.3.8. LetM be a model of T . Then there is an interpretation of theSkolem functions makingM a model of TSk.
Proof . We must give a meaning to the Skolem function symbols fϕ. Let ϕ(x, y)be an L-formula, and m ∈M have same length as y. IfM |= ∃x ϕ(x,m), thenthere is µ ∈M such thatM |= ϕ(µ,m); we choose such a µ and set fMϕ (m) = µ.If there is no suitable µ, we map m to any element in M .
With heavy use of the axiom of choice (we perform infinitely many choices),we thus define functions fMϕ , which are the interpretations of the Skolem func-tion symbols. By construction,M |= ∀y
(∃x ϕ(x, y) → ϕ(fϕ(y), y)
).
Let us return to our problem. If we now close a subset A ⊆ M under allfunctions including the Skolem functions, we are in good position to find anelementary inclusion, as all necessary witnesses have been plugged in.
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Theorem 1’.3.9 (descending Löwenheim-Skolem theorem). Let M be an infi-nite L-structure and A ⊆ M . Let κ be a cardinal with CardA + CardL ≤ κ ≤CardM . Then there isM0 �M such that A ⊆M0 and CardM0 = κ.
Proof . We may assume that CardA = κ. We turn M into an LSk-structurewhich satisfies TSk. In the language LSk, we now consider the substructure ofM generated by A (Lemma 1’.3.3). Forgetting the interpretation of the Skolemfunctions, this is a fortiori an L-structureM0. Moreover CardM0 = κ.
It remains to prove that the inclusion is elementary. We use Tarski’s Test(Theorem 1’.1.17). Let ϕ(m0) be a formula with parameters in M0; assumeM |= ∃x ϕ(x,m0). As M |= TSk, one has M |= ϕ(fMϕ (m0),m0). But byconstruction, the element fMϕ (m0) lies in M0. This is what we need in order toapply Tarski’s criterion.
Remark 1’.3.10.
• M0 � M as L-structures, not necessarily as LSk-structures. This isbecause when we expand the language to LSk, we create new formulas. Ifwe did want to ensureM0 �M as LSk-structures, we would need to addnew Skolem functions for the new formulas, getting LSk,Sk, etc.However an inclusion of L-structures is what we want; the expanded lan-guage LSk is a technical device we can forget about.
• The construction is highly non-canonical (it relies on the axiom of choice).In general, there is no smallest elementary substructure containing A:everything depends on the way we interpreted the Skolem functions, whichis arbitrary. One should not use the notation 〈A〉 for “Skolem hulls”.
As a final interesting word, we prove inconsistency of set theory.
Corollary 1’.3.11 (Skolem’s paradox). (If set theory is satisfiable) there is acountable model of set theory.
Proof . Write down ZFC as a first-order theory in the language {∈}. Start witha model and apply Theorem 1’.3.9.
However in this model U0 of set theory we may construct N and P (N),which is notoriously uncountable, and yet a subset of the countable underlyinguniverse. . .
[Catch: “countable” does not bear the same meaning in the first model, sayU , and the countable one U0. All that we have proved is that U0 is so small thatit does not “see” (i.e., perceive as a set) a bijection between P (N) and N.]
1’.3.3 The General Version and the Łoś-Vaught CriterionWe now try to “go up”; that is, provide very large elementary extensions ofexisting models.
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Theorem 1’.3.12 (ascending Löwenheim-Skolem theorem). LetM be an infi-nite L-structure and κ ≥ CardM be a cardinal. Then there is M∗ � M suchthat CardM∗ ≥ κ.
Proof . Let C′ = {ci : i < κ} be new constants, and consider the languageL′ = LM ∪ C′. Form the L′-theory T ′ = Th(M,M) ∪ {ci 6= cj : i 6= j}. Thistheory is consistent: a finite fragment mentions a finite number of propertiesϕ(m), which are trivially satisfiable in M, and a finite number of formulasci 6= cj , meaning that there are at least n distinct elements: this is satisfiableinM again.
Hence any finite fragment of T ′ is satisfiable inM; by compactness, there is aa modelM∗ of T ′ (notice thatM∗ is not likely to beM). AsM∗ |= Th(M,M),one has M �M∗; on the other hand, the constants of C′ say that M∗ has atleast κ elements.
Combining Theorems 1’.3.9 and 1’.3.12, one derives a sharper version.
Theorem 1’.3.13 (Löwenheim-Skolem theorem). LetM be an L-structure andA ⊆M . Let κ be a cardinal with CardA+ CardL ≤ κ. Then there isM′ suchthat A ⊆M ′, CardM ′ = κ, andM′ �M orM′ �M.
Proof . Go up (Theorem 1’.3.12), then down (Theorem 1’.3.9).
Corollary 1’.3.14 (Łoś-Vaught test). Let T be a first-order theory in a languageL. If there is κ ≥ CardL such that T is κ-categorical, then T is complete.
Proof . LetM,N |= T . We aim at showingM≡ N .By the Löwenheim-Skolem theorem (Theorem 1’.3.13), there areM′ and N ′
of cardinal κ such thatM�M′ orM�M′, and similarly for N and N ′. Butin any caseM≡M′ and N ≡ N ′. In particular,M′,N ′ |= T .
AsM′ and N ′ are models of T of cardinal κ, they are isomorphic by cate-goricity; it followsM′ ≡ N ′. As a conclusion,M≡M′ ≡ N ′ ≡ N .
Example 1’.3.15. We shall prove later (Theorem 1’.4.4) that DLO is ℵ0-categorical. In particular it will be complete!
End of Lecture 22.
Lecture 23 (Back-and-Forth Methods)
1’.4 Back-and-Forth MethodsIt is not very common in practice to have an isomorphism; a very good rea-son is that two structures may have the same theory without having the samecardinality! So one should aim at weaker correspondences.
And precisely, we shall work with very partial functions, functions which aredefined only on a finite set. But these turn out to be enough (as our formulasare finite too!) Model theory has a wide variety of so-called back-and-forth
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methods which enable to inductively extend partial functions. One should bearin mind the image of a bootstrap.
Disclaimer: back-and-forth methods are easier to work out in the case ofrelational languages. With unary functions they are still manageable; if thereis a binary function around, it is not very likely that one will understand anykind of back-and-forth.
1’.4.1 Dense Linear OrderingsWe start with a key example.
Definition 1’.4.1 (DLO). A dense linear ordering (“with no endpoints” be-ing always implicit) is a model of the following theory in the language {<} oforderings:
• ∀x ¬(x < x)
• ∀x∀y (x < y ∨ x = y ∨ y < x)
• ∀x∀y∀z (x < y ∧ y < z → x < z)
• ∀x∀y∃z x < y → (x < z ∧ z < x)
• ∀x∃y∃z y < x ∧ x < z
DLO denotes either this theory, or a model of the theory (this is harmless asthe theory will turn out to be complete).
Remark 1’.4.2. The first three axioms express that < is a linear ordering; thefourth (density) says that between any two points there is another point; thefifth (no endpoints) states that there is no least, nor largest element). In spiteof the name, one should never forget the last one!
Example 1’.4.3. Recall that given two orderings A and B, A∐<B denotes
the extension of the orderings such that any element of A lies below any elementof B (“B is pasted after A”).
• (Q, <) is a countable DLO.
• (Q∐<Q, <) and (Q
∐<{∞}
∐<Q, <) are other countable DLO’s.
• (R, <) is a DLO of cardinality continuum.
A close inspection will reveal that Q∐<Q and Q are actually isomorphic.
Back-and-forth methods start here.
Theorem 1’.4.4 (Cantor). Any two countable DLO’s are isomorphic.
Proof . LetM,N be countable DLO’s. We shall inductively construct an iso-morphism. To this extent, we enumerate the underlying setsM = {mk : k ∈ N}and N = {nk : k ∈ N}. We want to construct an isomorphism; it suffices toconstruct an order-preserving function σ defined on all ofM , and which is onto.This will be done inductively by considering two kinds of steps:
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• at odd stages, we shall make sure that σ is defined everywhere on M ;
• at even stages, we shall make sure that σ reaches every element of N .
We thus want to find an increasing sequence of partial maps which preservethe ordering. Let us start by defining σ0(m0) = n0. So far, so good, as all wecan say is m0 = m0, which certainly holds of n0.
• (stage 2k+1) Suppose that σ2k is a partial map with finite support, andsuch that a = dom σ2k ∈M and b = im σ2k ∈ N satisfy the same relations.(You may think that σ2k “preserves the configuration of the tuple”, thatit is a partial L-morphism in the sense of Definition 1’.2.2.) We want toextend σ2k to an order-preserving map σ2k+1 with mk+1 ∈ dom σ2k+1.mk+1 is in a certain configuration with respect to the finite tuple a. Ifmk+1 lies above any element of a, then as there are no endpoints in N ,there is certainly β ∈ N which lies above any element of b. If on the otherhand, mk+1 lies between two elements of the tuple a, then as N is dense,the same pattern can be reproduced in N . In any case there is β ∈ N suchthat the extended tuples a,mk+1 and b, β still satisfy the same relations.We let σ2k+1 extend σ2k by σ2k+1(mk+1) = β.
• (stage 2k+ 2) Suppose that σ2k+1 has been constructed as a finite partialmap which preserves relation between a′ = dom σ2k+1 and b′ = im σ2k+1.We aim at extending σ2k+1 in such a way that nk+1 will be in the image.And we argue exactly like above, getting in any case an element α ∈ Msuch that α behaves with respect to a′ exactly the way nk+1 does for b′.We let σ2k+2(α) = nk+1.
At the limit, we let σ = ∪k∈Nσk. This is still an order-preserving bijection.Because of our construction at odd stages, every element ofM will appear in thedomain, so dom σ = M . Because of our construction at even stages, im σ = N .It follows that σ :M→N is an isomorphism.
Bearing in mind Definition 1’.2.14, one sees that Theorem 1’.4.4 says thatDLO is ℵ0-categorical. As a consequence, one can deduce from Corollary 1’.3.14that DLO is complete.
Remark 1’.4.5. This superb proof relies on two essential ideas.
• If you can always extend a partial function, then at the limit somethingnice will happen. Proposition 1’.4.14 below will generalize this idea.It is however difficult to understand at first in what measure exactly count-ability is needed. Many constructions in model theory rely on inductionbeyond ω. But here we know how to deal only with finite configurations,and this is why the limiting process can’t go further than the first infi-nite. For instance, there are two non-isomorphic DLO’s of cardinality ℵ1(Counter-example 1’.4.15 below).
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• Above all, the proof relies on this bootstrap construction, arguing thatany finite configuration on one hand can be reflected on the other hand.This is worth a general definition (Definition 1’.4.10 hereafter).
1’.4.2 ∞-isomorphismsDefinition 1’.4.6 (local isomorphism, 0-isomorphism).
• A local isomorphism, or 0-isomorphism, σ :M→N is an injective partialfunction M → N with finite domain which preserves the constants, func-tions, and relations. Hence σ is a local isomorphism iff dom σ and im σsatisfy the same atomic formulas (inM,N respectively).
• Two tuples a ∈M and b ∈ N are locally, or 0-isomorphic, written a '0 b,if there is a local isomorphism σ mapping a to b.
• Two structuresM and N are locally, or 0-isomorphic, writtenM '0 N ,if the empty function ∅ 7→ ∅ is a 0-isomorphism.
“Local” is of course, opposed to “global”: we consider only functions withfinite support. So a local isomorphism is a small fragment of a morphism. Whenone wants to insist on the poor quality of σ, one calls it a 0-isomorphism: itpreserves only formulas of quantification rank 0. This is very far from beingelementary!
Remark 1’.4.7. A beginner’s mistake is to think that the empty tuple of Mis always 0-isomorphic to the empty tuple from N . This is not the case, asthere might be constants in the language. For instance, the ring Z/2Z satifies1 + 1 = 0, but the ring Z/3Z doesn’t. So even the empty function is notcompatible with the interpretation!
A little thinking will reveal thatM'0 N iff Th0(M) = Th0(N ) (quantifier-free theories, no parameters: this boils down to atomic formulas using termsbuilt from constants).
Example 1’.4.8.
• Consider the orderings (Z, <) and (Z∐< Z, <); we denote by 0 the 0 of
Z; there are two copies of 0 in Z∐< Z, which we denote respectively by
01 and 02. Clearly ∅ '0 ∅. Clearly too, 0 '0 01. Also, (0, 2) '0 (01, 02)(as 0 < 2 and 01 < 02).
• Consider the successor function and the structures (Z, s) and (Z∐
Z, s).Clearly ∅ '0 ∅ and 0 '0 01. But it is no longer the case that (0, 2) '0(01, 02). Indeed,M |= 2 = s(s(0)), whereas N |= 02 6= s(s(01)).
Example 1’.4.9. A moment’s thought and one realizes that in the language oforderings, local isomorphisms are exactly order-preserving bijections with finitesupport.
We can now give the key definition. Bear in mind the image of a bootstrap.
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Definition 1’.4.10 (∞-isomorphism, Karp family). Let M and N be two L-structures. M and N are ∞-isomorphic, written M '∞ N , if there is a non-empty family F of local isomorphisms M → N such that:
forth: for all σ ∈ F and α ∈M , there is τ ∈ F extending σ such that α ∈ dom τ
back: for all σ ∈ F and β ∈M , there is τ ∈ F extending σ such that β ∈ im τ
F is called a Karp family, and its elements are called ∞-isomorphisms.
Remark 1’.4.11.
• If there is a Karp family, there is a largest one (take the union of all).
• M ' N impliesM'∞ N (take all restrictions of a global isomorphism),but the converse need not hold in general: perhapsM and N don’t evenhave the same cardinality (Counter-example 1’.4.13 below).
Proposition 1’.4.12. Any two DLO’s are∞-isomorphic. More specifically, thefamily of local isomorphisms is a Karp family.
Proof . Local isomorphisms between orderings are just order-preserving, finitesupport bijections (which do exist: the family is not empty!). The back-and-forth property is the core of the proof of Theorem 1’.4.4.
We emphasize that ∞-isomorphic is much weaker than isomorphic.
Counter-example 1’.4.13. Consider the orderings (Q, <) and (R, <). Theyare ∞-isomorphic, but of course Q 6' R.
There is however a partial correction.
Proposition 1’.4.14. IfM'∞ N and both are countable, thenM' N .
Proof . This is an abstract version of the bootstrap construction used in The-orem 1’.4.4.
Propositions 1’.4.12 and 1’.4.14 together yield the proof of Theorem 1’.4.4.
Counter-example 1’.4.15. The notions of ∞-isomorphism and of a Karpfamily (Definition 1’.4.10) are about arbitrarily large finite configurations; oneshould therefore not expect to generalize Proposition 1’.4.14 past ℵ0, even ifMand N have the same cardinality. Here is a counter-example.
Consider an ordered sumM =∐ℵ1
Q (copies of Q copied after each other),and N =
∐ℵ1
(Q ∪ {∞}) (Q followed by a point). Both are dense linear order-ings without endpoints, of cardinality ℵ1. They are ∞-isomorphic (Proposition1’.4.12), and yet they are not isomorphic. The latter is non-trivial.
Back-and forth arguments are especially powerful in ω-saturated models(defined below, Definition 1’.5.12). For instance, it is not true that any twoalgebraically closed fields are ∞-isomorphic, but any two ω-saturated models(defined below, §1’.5.2) are. These are algebraically closed fields of infinite tran-scendence degree; you may already try to establish a back-and-forth betweenthem. This will be done in §1’.5.3.
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1’.4.3 Finitary Back-and-Forth*We build on local isomorphisms to find weaker versions of back-and-forth.
Definition 1’.4.16 (n+ 1-isomorphism).
• A local isomorphism σ : a→ b is an (n+ 1)-isomorphism if:
for all α there is β and an n-isomorphism τ : aα 'n bβ extending σfor all β there is α and an n-isomorphism τ : aα 'n bβ extending σ
• a and b are (n+1)-isomorphic, written a 'n+1 b, if there exists an (n+1)-isomorphism between them.
• Two structuresM and N are (n+ 1)-isomorphic, writtenM 'n+1 N , ifthe empty function ∅ 7→ ∅ is an (n+ 1)-isomorphism.
Remark 1’.4.17. Notice that this is an inductive definition, as opposed tothe notion of ∞-isomorphism (Definition 1’.4.10), which is defined for a class,simultaneously.
Here begin the funny games: given two structures, how isomorphic are they?
Remark 1’.4.18. Back-and-forth may be viewed as a game opposing two play-ers, having two different boards M and N . A round consists of the followingtwo steps:
• player 1 puts a pawn on either board;
• player 2 puts a pawn on the other board, in order to keep both configura-tions similar.
M and N are n-isomorphic if (a clever) player 2 is sure to last at least n rounds.
Example 1’.4.19. Consider Example 1’.4.8 again.
• ConsiderM = (Z, s) and N = (Z∐
Z, s). They are 0-isomorphic (as thelanguage has no constants!). They are 1-isomorphic, as picking an elementin one structure can certainly be reflected on the other one.They are however not 2-isomorphic. Consider the element 01 ∈ N ; it willbe reflected inM by some element which we may assume to be 0; 0 '0 01.But now player 1 chooses 02, which cannot be reflected inM, as there isonly one orbit of s inM, and 01 and 02 lie in distinct orbits of N .Interestingly, these structures turn out to be elementary equivalent (thismight be a little unclear yet but §1’.5.2 and §1’.5.3 will shed some light).
• Consider (Z, <) and (Z∐< Z, <). Let k ∈ N; I say that Z and Z
∐< Z are
k-isomorphic. For clarity, let us call Z the Z which comes alones; and letZ1 and Z2 be the left and right Z’s of the pasted structure.
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Player 2 will identify Z with Z1: whenever player 1 makes a move in Z,player 2 does the same in Z1; whenever player 1 makes a move in Z1,player 2 does the same in Z. Now, when player 1 plays in Z2, player 1picks a similar configuration in very big elements of Z, of size around 2k.Player 1 will win, but it will take him at least k rounds!
There is a name for the second phenomenon of Example 1’.4.19.
Definition 1’.4.20 (ω-isomorphism).
• σ : a→ b is an ω-isomorphism if for all n it is an n-isomorphism.
• a et b are ω-isomorphic if there is an ω-isomorphism between them.
• Two structures are ω-isomorphic if ∅ is an ω-isomorphism.
Remark 1’.4.21.
• A beginner’s mistake: ω denotes the first infinite ordinal, which is verysmall for set theory. In particular, ω-isomorphism is way weaker than∞-isomorphism; here is why. In Example 1’.4.19 we have seen that (Z, <)and (Z
∐< Z, <) are ω-isomorphic. However they are not ∞-isomorphic,
as otherwise, being countable, they would be isomorphic by Proposition1’.4.14, which is clearly not the case.
• ω-isomorphisms are a little subtle. They mean that player 2 always has astrategy that will keep him playing for at least k rounds, but the choice ofthe strategy might depend on k (typically Example 1’.4.19). If there werea uniform strategy, which does not depend on how long player 2 tries tokeep playing, one would have ∞-isomorphism (Definition 1’.4.10 above).
ω-isomorphisms are especially interesting because of our notion of a lan-guage. Going down through the recursive definition of satisfaction, one seesthat our formulas are always about finite tuples, of arbitrary length.
Proposition 1’.4.22. IfM'ω N , thenM≡ N .
In order to prove this proposition, let us introduce useful terminology.
Definition 1’.4.23 (type of a tuple). Let M be an L-structure and a ∈ Mbe a tuple (abusing, as usual, Cartesian powers). The type of a in M istpM(a) = {ϕ(x) : M |= ϕ(a)}. When there is no risk of confusion, one dropsthe superscript M.
Bearing in mind the definition of the quantification rank of a formula (Defini-tion 1’.0.8), one can even define tpMn (a) = {ϕ(x) : qrk(ϕ) ≤ n andM |= ϕ(a)}.
Lemma 1’.4.24. Let M, N be L-structures and a, b be tuples extracted fromM, N respectively. If a 'n b, then tpn(a) = tpn(b).
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Proof . By induction. First notice that by definition (Definition 1’.4.6), if a '0b, then a and b satisfy the same relations, hence the same atomic formulas; theclaim is proved.
Suppose that the property is known for n; we show it for n + 1. So leta 'n+1 b be two n+ 1-isomorphic tuples. We show tpn+1(a) = tpn+1(b). Let ϕbe a formula of tpn+1(a). We show that N |= ϕ(b).
A quick induction on the structure of ϕ reveals that the only interesting caseis that of ϕ(a) being ∃x ψ(a, x), for a formula ψ now of quantification rank ≤ n.
We have assumedM |= ϕ(a): so there exists α ∈M such thatM |= ψ(a, α).In particular, ψ ∈ tpn(a, α). As a and b are n + 1-isomorphic, there is bydefinition (Definition 1’.4.16) β ∈ N such that a, α 'n b, β. By induction,ψ ∈ tpn(b, β); that is, N |= ψ(b, β). It follows that N |= ϕ(b): we are done.
Remark 1’.4.25. We insist that if a and b are locally isomorphic, and thatthere is a Karp family of extensions of a '0 b (one may then write a '∞ b), aand b satisfy the same formulas with quantifiers.
Proof of Proposition 1’.4.22. Suppose M 'ω N ; we show M ≡ N . Ourassumption means that for each n, ∅M 'n ∅N . By Lemma 1’.4.24, tpMn (∅) =tpNn (∅). But tpMn (∅) = Thn(M) (the theory of M restricted to formulas ofqrk ≤ n). As this is true for any n, one gets Th(M) = Th(N ); soM≡ N .
Proposition 1’.4.22 does not have a converse, because when n > 0, Lemma1’.4.24 doesn’t.
Counter-example 1’.4.26. Consider the case of (Z, s) and (Z∐
Z, s) again.Then 0 and 01 clearly satisfy the same formulas of qrk ≤ 1, but as observed,they are not 1-isomorphic.
Remark 1’.4.27. The choice of Counter-example 1’.4.26 is typical. Actually, ina language consisting of finitely many relations, the converse to Lemma 1’.4.24does hold, and in particularM'ω N iffM≡ N !
We sum up the isomorphism strength as follows (n < m)
' ⇒ '∞ ⇒ 'ω ⇒ 'm ⇒ 'n⇓≡
Counter-example 1’.4.28.
• (R, <) '∞ (Q, <) but they are not isomorphic (Counter-example 1’.4.13)
• (Z, <) 'ω (Z∐< Z, <) but they are not ∞-isomorphic (Remark 1’.4.21).
• (Z, s) ≡ (Z∐
Z, s) but they are not ω-isomorphic (Example 1’.4.19).
• (Z, s) '1 (Z∐
Z, s) but they are not 2-isomorphic (Example 1’.4.19).
In special cases, converses do hold:
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• IfM'∞ N are countable, thenM' N (Proposition 1’.4.14).
• There always is an ordinal such thatM'α N impliesM'∞ N .
• In a finite, purely relational language,M≡ N does imply N 'ω N .
End of Lecture 23.
Lecture 24 (Elimination Theorem)
1’.5 Quantifier Elimination and ω-SaturationSometimes, every formula (with free variables!) boils down to a formula withno quantifiers. This was typically the case when in high school, you realizedthat ∃x ax2 + bx+ c = 0 iff b2 − 4ac ≥ 0 (over Th(R) as an ordered ring). Wegive a name to this general phenomenon.
Definition 1’.5.1 (quantifier elimination). A first-order theory T eliminatesquantifiers (“has QE”) if for any formula ϕ(x) there is a quantifier-free formulaψ(x) such that T |= ∀x ϕ(x)↔ ψ(x).
(One says that ϕ(x) and ψ(x) are equivalent modulo T ; this amounts tohaving, in L ∪ {x}, T ∪ {ϕ(x)} |= ψ(x) and T ∪ {ψ(x)} |= ϕ(x).)
Example 1’.5.2.
• In R as an ordered ring, the formula ∃x ax2 + bx+ c = 0 is equivalent to(b2 − 4ac ≥ 0) ∨ (a = 0 ∧ (b 6= 0 ∨ c = 0)). Actually the theory RCF of Ras an ordered ring eliminates quantifiers.
• In C as a ring, the formula ∃x anxn + · · · + a1x + a0 is equivalent to
¬(∧ni=1ai = 0 ∧ a0 6= 0) (this is the D’Alembert-Gauß theorem). Actuallythe theory ACF0 of C as a ring eliminates quantifiers.
We shall give a semantic criterion for a theory to admit QE (Corollary 1’.5.6below). As it can be stated in a more general form, we extend the definition toarbitrary elimination.
1’.5.1 Elimination SetsDefinition 1’.5.3 (elimination set). Let T be a theory. Let E be a set offormulas. E is an elimination set modulo T if every formula ϕ of L is equivalentmodulo T to some Boolean combination of formulas from E.
Theorem 1’.5.4 (elimination theorem). Let T be a theory and E a set of for-mulas. Then E is an elimination set if and only if, for all modelsM and N ofT and all tuples m and n extracted from M and N respectively,
if m and n satisfy the same formulas from E, then they satisfy thesame formulas.
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Proof . One implication is trivial. For the other one, we may clearly assumethat E is closed under Boolean operations. (In particular, E contains a con-tradiction.) Let ϕ(x) be an L-formula. We shall prove that ϕ(x) is equivalent(modulo T ) to a formula of E.
Step 1. There is a finite subset Ψ ⊆ E such that the theory T ∪{ψ(c)↔ ψ(d) :ψ ∈ Ψ} ∪ {ϕ(c) ∧ ¬ϕ(d)} is inconsistent.
Verification: Let c, d be tuples of new constant symbols (of the same length asx) and L′ = L ∪ {c, d}. Let
T ′ = T ∪ {ϕ(c) ∧ ¬ϕ(d)} ∪ {ψ(c)↔ ψ(d) : ψ ∈ E}
The L′-theory T is not consistent; supppose it is. Then there is a modelM |= T ′
with interpretations m = cM and n = dM. Notice that m and n satisfy thesame formulas from E, but disagree on ϕ: against the assumption. So T ′ is notconsistent. By compactness, there is a finite, inconsistent fragment. ♦
Step 2. For everyM |= T , there is a formula ψM(x) ∈ E equivalent to ϕ(x).
Verification: Let M |= T , and X = {m ∈ M : M |= ϕ(m)} (the subset ofMn defined by ϕ). If X is empty, then bearing in mind that E contains acontradiction, we are done. So we may suppose that X 6= ∅.
For each m ∈ X, let Ψm = {ψ ∈ Ψ : M |= ψ(m)} and Θm = Ψ \ Ψm. Letψm =
(∧Ψmψ
)∧(∧Θm¬ψ
)(hence ψm describes exactly which formulas of Ψ
a tuple m ∈ X satisfies).Notice that each ψm is in E, and that only finitely many ψm’s are possible
by finiteness of Ψ (even though X might be infinite). Let ψM = ∨Xψm: thisshould be another definition of X! Notice that ψM ∈ E. We now claim thatM |= ∀x (ϕ(x)↔ ψM(x)).
Let n ∈ M . If M |= ϕ(n), then by definition n ∈ X; hence M |= ψn(n)and M |= ψM(n). On the other hand, assume M |= ψM(n). By constructionof ψM, there is m ∈ X such thatM |= ψm(n); by construction of ψm, one hasM |= ψm(m). It follows that m and n satisfy exactly the same formulas fromΨ. By Step 1, m and n must agree on ϕ too. As m ∈ X, one hasM |= ϕ(m);it followsM |= ϕ(n). ♦
The issue is that, so far, the formula ψM is equivalent to ϕ only inM (if thetheory T were complete we would however be done). We solve this problem.
Step 3. Let M |= T . There is a sentence θM ∈ E true in M and such thatT ∪ {θM} |= ∀x (ϕ(x)↔ ψM(x)).
Verification: Let ΘM = {χ ∈ E :M |= χ} (this is E ∩ Th(M)) and
T ′ = {χ : χ ∈ ΘM} ∪ {¬χ : χ 6∈ ΘM} ∪ {¬∀x (ϕ(x)↔ ψM(x))}
If T ′ were consistent, we would have a model N of T ′. The empty tuple ∅ ∈N satisfies exactly the same formulas from E as ∅ ∈ M ; in particular, by
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assumption, ∅ satisfies the same formulas in M and N , that is M ≡ N . YetM |= ∀x (ϕ(x)↔ ψM(x)), a contradiction. So T ′ is not consistent.
By compactness, there is a finite fragment of ΘM, which we can take to bea single sentence θM, such that T ∪ {θM} |= ∀x (ϕ(x)↔ ψM(x)). ♦
Step 4. Finitely many θM’s suffice for all models of T .
Verification: Let T ′ = T ∪ {¬θM : M |= T} (this makes sense: the collectionof models is a proper class, but the collection of θM’s is a set alright, as asubcollection of the set of L-sentences). If T ′ were consistent, we would have amodel N |= T in which none of the θM’s holds, which is impossible!
By compactness, T ′ is finitely inconsistent, and this means that there is afinite set Θ of θM’s such that T |= ∨ΘθM. ♦
Step 5. There is ψ(x) ∈ E such that T |= ∀x ϕ(x)↔ ψ(x).
Verification: Let Θ = {θM :M |= T}; we may assume that Θ is finite. Consider
ψ(x) =∨Θ
(θM ∧ ψM(x))
Notice that ψ ∈ E. Now letM |= T , andm ∈M : we showM |= ϕ(m)↔ ψ(m).Suppose M |= ϕ(m). By construction, M |= θM. In particular, M |=
∀x (ϕ(x) ↔ ψM(x)), whence M |= ϕ(m) ↔ ψM(m), and M |= ψM(m),whenceM |= ψ(m).
Suppose now M |= ψ(m). Then there is a model N such that M |= θN ∧ψN (m). As T ∪ {θN } |= ϕ(c)↔ ψN (c), we see thatM |= ϕ(m). ♦
This concludes the proof of Theorem 1’.5.4.
Remark 1’.5.5. This proof is a little long and clumsy. We have just beencovering a normal, compact space; a topological argument is possible, short,and elegant, but it requires introducing spaces of types, which is a little beyondthe scope of a first introduction.
Exercise 1’.9
Corollary 1’.5.6. A theory T eliminates quantifiers if and only if, for all modelsM and N of T and all tuples m and n extracted from M and N respectively,
if m and n satisfy the same atomic formulas, then they satisfy thesame formulas.
The question is now: how do we prove that two tuples do satisfy the sameformulas? In practice, this often results from a back-and-forth construction.
Corollary 1’.5.7. Let T be a first-order theory such that for any two modelsM,N |= T , local isomorphisms between M and N form a Karp family. ThenT eliminates quantifiers.
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Proof . We use Corollary 1’.5.6. Let m,n be drawn from M,N respectivelyand satisfy the same quantifier-free formulas; we show that they satisfy thesame formulas. As m and n satisfy the same quantifier-free formulas, they arelocally isomorphic. By assumption, they are ∞-isomorphic. So m '∞ n, andit follows tpM(m) = tpN (n) (see Remark 1’.4.25). Hence m and n satisfy thesame formulas; T eliminates quantifiers.
We shall later generalize Corollary 1’.5.7 to a weaker assumption (Theorem1’.5.19 below).
Remark 1’.5.8. QE is strongly language-dependent; actually for any L-theoryT , there is a language L ⊆ L′ and an L′-theory T ⊆ T ′ such that T ′ eliminatesquantifiers.
Verification: We first get rid of ∀’s, replacing them by ¬∃¬. For any formula∃y ϕ(x, y) we introduce a new relation symbol Rϕ(x) and we add to the theory∀x (∃y ϕ(x, y))↔ Rϕ(x).
Of course the resulting theory T1 in the resulting language L1 need noteliminate quantifiers, as we have created new formulas. But we repeat theconstruction. After ω steps, any formula of L′ = ∪nLn is equivalent moduloT ′ = ∪nTn to a quantifier-free formula of L′. ♦
One can actually prove compactness that way, reducing to compactness ofpropositional logic.
End of Lecture 24.
Lecture 25 (ω-Saturated Models)
1’.5.2 ω-Saturated ModelsTheorem 1’.5.4 suggests a method to prove quantifier elimination: just show thattwo locally isomorphic tuples satisfy the same formulas. This can be done byestablishing ∞-back-and-forth (ω-back-and-forth is weaker and sufficient butmore subtle: hope for the strongest.) So let us go back to back-and-forthconstructions.
For instance, if one tries to show that Th(C) eliminates quantifiers, oneshould start with algebraically closed fields K,L; one would take two locallyisomorphic tuples a '0 b, and try to check that we can always extend. Thisfails if, for instance, K has an element which is transcendental over a, but L isalgebraic over b.
The conclusion of this algebraic discussion is that some models are “richer”than others, and that presumably back-and-forth is easier to manage in richmodels. A formal notion of wealth is ω-saturation (Definition 1’.5.12 below).This requires the following definition, which generalizes the notion of a systemof equations in n variables. To understand it, fix an integer n and a tuple ofvariables x = (x1, . . . , xn) of length n.
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Definition 1’.5.9 (type). Let L be a first-order language, M an L-structure,A ⊆M a set of parameters. An n-type over A is a collection p(x) of LA-formulasϕ(x, a) (a ∈ A) which is consistent with Th(M, A).
Treating the variables as constants, one sees that an n-type is merely atheory in LA(x) extending Th(M, A).
This generalizes the case of the type of a tuple (Definition 1’.4.23), whichis called realized. But any type is the type of a tuple which “hasn’t appearedyet”: every type is realized somewhere.
Remark 1’.5.10. If p(x) is a type over A ⊆ M , then there is N � M andn ∈ N such that n realizes p. (Of course N will tend to be a proper extensionofM.)
Types are amazingly important and interesting (they come with nice topo-logical methods), and students should refrain the teacher from going to deepinto this matter.
Example 1’.5.11.
• If A ⊆M and b ∈M , then tp(b/A) = {ϕ(x, a) : ϕ ∈ LA :M |= ϕ(b, a)} isan n-type over A.
• Consider (N, <). Then {x > n : n ∈ N} is a 1-type over N. Notice that itis not realized in N, but in some elementary extension.
• When we constructed the non-standard reals in §1.6.3, we simply con-structed a 1-type over R, namely p(x) = {0 < x < r : r ∈ R>0}, and thenwent to an elementary extension R∗ � R satisfying it.
• Consider C as a ring, and study maximal 1-types over Z. There are twokinds:
– algebraic types, types which contain a formula P (x) = 0 for a poly-nomial P with coefficients in Z.
– the transcendental type, which contains {P (x) 6= 0 : P ∈ Z[X]}.
• We build on the previous example to study (2-)types of pairs of complexnumbers p(x, y) over Z. There are five kinds:
– both x and y are algebraic over Z: the type boils down to {P1(x) =0, P2(y) = 0} for some polynomials P1 and P2.
– x is algebraic over Z, but y is transcendental: the type reduces to{P1(x) = 0} ∪ {P (y) 6= 0 : P ∈ Z[X]} for a certain polynomial P1.
– x is transcendental, y is algebraic: p(x, y) = {P (x) 6= 0 : P ∈ Z[X]}∪{P2(y) = 0} for some P2.
– x and y are transcendental but the family (x, y) is algebraically de-pendent: the type contains {P (x) 6= 0 : P ∈ Z[X]}∪{P (y) 6= 0 : P ∈Z[X]} ∪ {Q1(x, y) = 0} for some Q1 ∈ Z[X,Y ].
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– The family (x, y) is algebraically independent: the type is p(x, y) ={Q(x, y) 6= 0 : Q ∈ Z[X,Y ]}. This is the type of any pair of tran-scendence degree 2.
One measures richness of a structure by the amount of types it realizes.
Definition 1’.5.12 (ω-saturated). An L-structure M is ω-saturated if any 1-type over finitely many parameters is realized inM.
Example 1’.5.13.
• A finite structure is always ω-saturated, but this is a trivial example!
• Any DLO is ω-saturated.
• (N, s) is not ω-saturated. Let ϕn(x) be the formula x 6= sn(0). Then the1-type p(x) = {ϕn(x) : n ∈ N} uses 0 as its only parameter, and is notrealized in N.
• (N, <) is not ω-saturated. Let ψn(x) be the formula ∃x1 . . . ∃xn x1 < · · · <xn < x. Then the 1-type {ψn(x) : n ∈ N} uses no parameters and is notrealized in N.
• The ring Q is not ω-saturated. Indeed, any element of Z can be written asa sum of 1’s. So the transcendental type p = {P (x) 6= 0 : P (X) ∈ Z[X]}uses no parameters. It is however not realized in Q.
Theorem 1’.5.14. LetM be an infinite L-structure. Then there is an elemen-tary extensionM�M∗ which is ω-saturated.
Proof . This proof relies on ordinal induction, and may be omitted freely; theresult may not.
We enumerate all finite tuples a of elements of M , and all 1-types over a,finding an enumeration (by some ordinal λ perhaps bigger than ω) of all 1-types over all finite tuples of M : {pi(x) : i < λ}. The reader not familiar withordinals ought to suppose λ = ω for simplicity; in this case the construction isvery understandable.
LetM0 =M. For each i, pi(x) is a 1-type overMi, so it is realized in someelementary extensionMi+1 ofMi (a quick look at Remark 1’.5.10 if necessary).Take unions at limit ordinals (this is required only if λ > ω). Let M1 = Mλ.As we have taken the increasing union of an elementary chain, Lemma 1’.1.16says that M � M1. By construction, all 1-types over finite tuples of M arerealized inM1.M1 need not be ω-saturated, as we have added new elements, so we might
have created new types. But we resume the construction, getting M2. Aftercountably many steps, the unionM∗ is ω-saturated, and an elementary exten-sion ofM by Lemma 1’.1.16 again. e:ultraproductssaturated
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Remark 1’.5.15. One has no control on the cardinality of an ω-saturated ele-mentary extensionM∗, even starting with a countable base modelM. Indeed,at each step the number of types might be huge.
For instance, if one considers Z as a ring, then there are continuum manydifferent 1-types: for each subset I of the primes, consider pI = {x 6= 0} ∪{∃y x = p.y : p ∈ I}∪{∀y x 6= p.y : p 6∈ I} which expresses that the only primesdividing x are the elements of I.
So there is no countable, ω-saturated elementary extension of the ring Z.
Dealing with 1-types or n-types is the same, as a quick induction reveals.
Lemma 1’.5.16. M is ω-saturated iff any n-type over finitely many parametersis realized inM.
Proof . Induction. Suppose any n-type with finitely many parameters is real-ized inM, and let p(x, y) be an n+ 1-type with finitely many parameters a (xstill stands for (x1, . . . , xn); the n+1th variable is y).
Let q(x) = {∃y ϕ(x, y, a) : ϕ(x, y, a) ∈ p}. This is clearly an n-type over a;by induction, it is realized by an n-uple b ∈M .
We now let r(y) = {ϕ(b, y, a) : ϕ(x, y, a) ∈ p}. This is clearly a 1-type overb, a, which is again a finite tuple. So r is realized inM: there is c ∈M realizingit. It is clear that b, c realizes p.
Remark 1’.5.17. Observe that one couldn’t have taken both projections of p,constructing and realizing first q, then s(y) = {∃x ϕ(x, y, a) : ϕ(x, y, a) ∈ p}.Indeed, even though c is a realization of s, (a, c) has no reason to realize p!
Again, why are ω-saturated models so important? Because when one tries tocheck completeness of a theory, one may focus on them, which are presumablynicer than the general models. Here is a generalization of Lemma 1’.1.6 toweaker assumptions.
Lemma 1’.5.18. T is complete iff any two ω-saturated models are elementaryequivalent.
Proof . LetM,N be models of T ; we show that they are elementarily equiva-lent. By Theorem 1’.5.14, there areM∗ �M, N ∗ � N which are ω-saturated.AsM≡M∗ and N ≡ N ∗,M∗ and N ∗ are models of T . By assumption, theyare elementary equivalent. It followsM≡M∗ ≡ N ∗ ≡ N ; we are done.
Here is our generalization of Corollary 1’.5.7.
Theorem 1’.5.19. Let T be a first-order theory such that for any ω-saturatedmodelsM,N |= T , local isomorphisms betweenM and N form a Karp family.Then T eliminates quantifiers.
Proof . We use the elimination theorem, Theorem 1’.5.4. Let M,N |= T ; welet m ∈ M , n ∈ N satisfy same quantifier-free formulas; we want to show thatthey satisfy the same formulas.
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First, we may assume thatM and N are ω-saturated. By Theorem 1’.5.14,there are ω-saturated elementary extensions M � M∗ and N � N ∗. As msatisfies the same formulas in M and M∗, it actually suffices to work in M∗and N ∗, that is we may assumeM and N are ω-saturated.
As m and n satisfy the same quantifier-free formulas, they are locally iso-morphic, so by assumption they are∞-isomorphic. They must therefore satisfythe same formulas. Exercise 1’.10
The conclusion of this abstract discussion is that ω-saturated elementaryextensions always exist, and form a good ground on which to study a theory(the details are not too important).
End of Lecture 25.
Lecture 26 (Examples)
1’.5.3 ExamplesWe eventually shed light on all notions by studying examples.
DLO’s
The theory of dense linear ordering has already been defined and studied in§1’.4.1. We put the existing pieces together.
Proposition 1’.5.20 (DLO’s). DLO is complete and decidable; it eliminatesquantifiers. The theory is ℵ0-categorical but not ℵ1-categorical. Moreover, anymodel is ω-saturated and local isomorphisms are ∞-isomorphisms.
Proof . We know everything, but here is how a study could go. Past the twofirst, many steps are permutable.
(i). The first step is to realize that any DLO is indeed ω-saturated.
(ii). Then, prove that any local isomorphism is an ∞-isomorphism.
(iii). Moreover, as there are no constants, ∅ is a local isomorphism. It followsfrom (ii) that any two models are ∞-isomorphic.
(iv). As a consequence of (iii) and with Proposition 1’.4.14, any two countablemodels are isomorphic; the theory is ℵ0-categorical.
(v). One may uses the Łoś-Vaught criterion (Corollary 1’.3.14 to argue thatthe theory is complete.
(vi). It would be as simple to say that any two models are elementary equivalentby (iii); so the theory is complete, again.
(vii). As the theory is clearly effectively presented (finitely many axioms!), andcomplete, it is decidable (Corollary 1.6.3).
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(viii). Turning back to categoricity questions, one would see that the theoryis not ℵ1-categorical (Counter-example 1’.4.15). It follows from Morley’stheorem (Theorem 1’.2.16) that it is not κ-categorical in any κ ≥ ℵ1.
(ix). It follows from (ii) that DLO eliminates quantifiers (Theorem 1’.5.19).
The Successor Function
Proposition 1’.5.21 (N, 0, s). Let T have the following axioms:
• ∀x∀x′ s(x) = s(x′)→ x = x′;
• ∀x s(x) 6= 0;
• ∀x∃y x 6= 0→ s(y) = x;
• ∀x sn(x) 6= x for each n ∈ N.
The theory is complete and decidable; it eliminates quantifiers. It is not ℵ0-categorical but is κ-categorical in any κ ≥ ℵ1. Moreover N embeds elementarilyinto every model.
Proof . This is a new study. We must understand the ω-saturated models. Inorder to do this we must develop some understanding of the general model.
(i). A model is a set equipped with a function s which is onto except on 0, andwhich has no cycles. There are therefore two kinds of orbits: the orbit of0, which starts and never ends, and other orbits, which have no startingnor ending point: they are like (Z, s).We see that the general model of T has the formMi = N
∐∐I Z, where
I is any set. There is no way to order Z-like orbits.
(ii). Local isomorphisms are easily understood; one would however fail to es-tablish back-and-forth between any two models. It is clear that N andN∐
Z are not ∞-isomorphic (being countable they would be isomorphicby Proposition 1’.4.14).
(iii). In particular, the theory is not ℵ0-categorical. On the other hand sinceisomorphism classes are determined by the sole set I, which bears nostructure, T is obviously κ-categorical in any uncountable κ.
(iv). One may apply the Łoś-Vaught criterion to derive completeness of T (andits decidability, as it is effectively axiomatized).However we insist that the sooner one understands the ω-saturated models,the better.
(v). The 1-type p(x) = {sn(0) 6= x : n ∈ N} (no parameters! N appears asindexing set!) is clearly not realized in N, but it is in N
∐Z. On the other
hand, if we fix a ∈ Z, then the 1-type q(x) = {sn(a) 6= x : n ∈ N} over a is
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not realized in N∐
Z. If we want to realize it, we must go to N∐
Z∐
Z′.And so on.We then understand that a model is ω-saturated iff of the form N
∐∐I Z
for an infinite set I.
(vi). One then sees that local isomorphisms between ω-saturated models doform a Karp family: suppose that a '0 b are drawn from ω-saturatedmodels, and let α be on the same side as a. We want to find β such thata, α '0 b, β.
• If α lies in the orbit of some ai, say α = sn(ai) for n ∈ N, we letβ = sn(bi). As a '0 b, this creates no confusion!
• If on the other hand α lies in a new orbit, we pick β in a new orbittoo: this is possible, as I is infinite!
(vii). Any two models are clearly locally isomorphic (the orbit of 0 is similarin either model). It follows from (vi) that any two ω-saturated modelsare ∞-isomorphic. It follows that the theory is complete and eliminatesquantifiers.
(viii). In particular, any inclusion of structures is elementary. As N embeds intoany model, it embeds elementarily into any model.
The Ordering
Proposition 1’.5.22 (N, 0, <). Let T be the theory:
• ∀x ¬(x < x)
• ∀x∀y (x < y ∨ x = y ∨ y < x)
• ∀x∀y∀z x < y ∧ y < z → x < z
• (∀y 0 ≤ y) ∧ (∀x∀y x ≤ y → x = 0)
• ∀x∃y∀z y > x ∧ (x < z → y ≤ z)(There is an element lying immediately above x)
• ∀x∃y∀z x 6= 0→ (y < x ∧ (z < x→ z ≤ y))(If x is not 0, then there is an element lying immediately below x).
The theory is complete and N embeds elementarily into any model. It does notadmit QE, but it does eliminate quantifiers in the language (0, <, {dn : n ∈ N}),where dn is a new binary predicate symbol standing for
dn(x, y) : ∃x1 . . . ∃xn−1 x < x1 < · · · < xn−1 < y
(”the algebraic distance from x to y is at least n”)
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Proof . Things are a little more subtle here.
(i). A key observation is that this theory interprets (N, 0, s): one can define (bya first-order formula) the successor function s. In particular, any modelof T has the formMI = N
∐<
∐(I,<) Z, where I is now an ordered set.
(ii). One must be very careful that local isomorphisms do not form a Karpfamily between ω-saturated models. Indeed, pairs (x, x+) and (x, x++)are locally isomorphic, but back-and-forth between them is not possible!We already have a feeling that distance comes into play. But we reservelanguage-related issues for the end.
(iii). If one has studied ω-isomorphisms (§1’.4.3), one sees that any two modelsare ω-isomorphic, so the theory is complete. There is another proof in (v)but this one is recommended, being simple and elegant.
(iv). For a model to be ω-saturated, I must be infinite. This is howevernot enough. Fix a in some copy of Z, and consider the 1-type p(x) ={∃x1 . . . ∃xn−1 a < x1 · · · < xn−1 < x : n ∈ N} over a. It means that xis infinitely bigger than a, that x lies in a copy of Z lying above that inwhich a lives.In particular, we understand that for the model to be ω-saturated, I musthave no largest element. And similarly, it must have no least element: ithas no endpoints. This is still not enough.Fix a < b in I, and consider the 1-type q(x) = {∃x1 . . . ∃xn−1 a < x1 · · · <xn−1 < x : n ∈ N} ∪ {∃x1 . . . ∃xn−1 x < x1 · · · < xn−1 < b : n ∈ N} over(a, b): it expresses that the orbit of x contains neither a nor b, that x livesin a copy of Z which lies between that of a and that of b. This is densityof the ordering I!And we now see thatMI is ω-saturated iff I is a DLO.
(v). One could reprove completeness: the theory is of course not ℵ0-categorical(neither is it ℵ1-categorical, since DLO isn’t), but any two countable, ω-saturated models are isomorphic.Indeed, let MI and MJ be such models. I and J are countable DLO’s,whence isomorphic. We then construct an isomorphism MI ' MJ byidentifying copies of Z.In particular, any two countable and ω-saturated models are elementaryequivalent. This is however not enough. Let N be any model; by theLöwenheim-Skolem theorem (Theorem 1’.3.13), it contains a countableelementary substructure N0 =MI , where I is countable. Now I embedsinto a countable DLO I ′, so N0 embeds elementarily into N1 = MI′ . Itfollows that N ' N1, and completeness is proved.This method is however not recommended, as it relies on the existence ofcountable, ω-saturated models; one should prefer the proof of (iii).
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(vi). We let L = {0, <} and L′ = L ∪ {dn : n ∈ N} be the expanded language.Notice that WFF(L) = WFF(L′); we don’t change the expressive strengthof the language, but merely force certain quantified formulas to now bequantfier-free.In particular, a model of T is ω-saturated in L iff ω-saturated in L′: ourdescription of ω-saturation does not change.
(vii). One now sees that L′-local isomorphisms between ω-saturated models areL′-∞-isomorphisms.From there, quantifier elimination in L′ is immediate. As N ⊆MI is alsoan inclusion of L′-structures, one finds N �MI .
Algebraically Closed Fields*
Definition 1’.5.23 (ACF). An algebraically closed field is a model of the fol-lowing theory in the language {0, 1,+,−, ·} of rings:
• the structure is a field;
• for each n ∈ N, the axiom ∀an . . . ∀a0∃x an 6= 0→ an · xn + · · ·+ a0 = 0.
ACF denotes either this theory, or a model of the theory.
We let P denote the set of prime numbers.
Definition 1’.5.24 (ACFq).
• For p ∈ P, an algebraically closed field of characteristic p is a model ofthe theory ACF∪{1 + · · ·+ 1︸ ︷︷ ︸
p times= 0}
• An algebraically closed field of characteristic 0 is a model of the theoryACF∪{1 + · · ·+ 1 6= 0 : p ∈ P}.
Proposition 1’.5.25. ACF eliminates quantifiers. Moreover, for any q ∈ P ∪{0}, ACFq is complete. It is not ℵ0-categorical, but is κ-categorical in anyuncountable κ.
Proof . We begin with simple remarks: given algebraically closed fields K,L,one has K '0 L iff they have the same characteristic. Moreover, an ACF isω-saturated iff it has infinite transcendence degree over its prime field.
It is then very easy to see that if K,L are ω-saturated ACF’s of the samecharacteristic, then local isomorphisms form a Karp family. In particular, K ≡L. ACFq is complete! Now the characteristic is in the quantifier-free type ofany tuple, so ACF eliminates quantifiers.
As for categoricity, Q and Q(X) are clearly non-isomorphic; but playingwith transcendence bases, one easily finds that ACFq is κ-categorical in anyuncountable κ.
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Corollary 1’.5.26 (Hilbert’s Nullstellensatz). Let K be an algebraically closedfield, X = (X1, . . . , Xn) a tuple of indeterminates, and I C K[X] an ideal ofK[X]. Then I has a solution in Kn.
Proof . We may assume that I is a maximal ideal m; notice that by Noetherian-ity, m is finitely generated. In particular, existence of a solution is a first-orderformula with parameters in K.
We let L be the algebraic closure of the field K[X]/m; clearly K ⊆ L. Asboth are models of ACF and the theory eliminates quantifiers, one actually hasK � L. Now there is a solution to m in Ln (the class modulo m of the tuple X);since this is first-order, there is a solution in Kn too.
More interesting results hold, but one unfortunately has to stop some day.
End of Lecture 26.
1’.6 ExercisesExercise 1’.1. Show Lemma 1’.1.16, that is show that given an increasingelementary chain (Mi)i∈I of L-structures, the union M∗ is an elementary ex-tension of eachMi.
Exercise 1’.2. Let T be a theory which has models of cardinality n for arbi-trarily large n ∈ N. Show that T has an infinite model.
Exercise 1’.3. Finish the proof of Lemma 1’.1.11 by proving directly that when-everM⊆ N and ϕ(m) is a universal formula with parameters m in M , one hasN |= ϕ(m)⇒M |= ϕ(m).
Exercise 1’.4. Let L = {R}, a binary relation, and T express that R is anequivalence relation.
1. LetM,N |= T . Find a necessary and sufficient condition forM |= N .
2. LetM⊆ N |= T . Find a necessary and sufficient condition forM� N .
No justification is required.
Exercise 1’.5. Let H,G be groups with H � G (the language contains Lgrp =(e,−1 , ·), but could be larger).
1. Prove that if G is simple, so is H.
2. Prove that the converse may fail.
Exercise 1’.6. A theory T is universal if it the set of consquences of a set ofuniversal sentences.
Also, if T is a theory, let T∀ be the set of all its consequences that areuniversal; that is
T∀ = {ϕ an L-sentence such that T |= ϕ}
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1. The language is {·}. Is the theory of groups universal? (no justificationrequired)
2. The language is {e,−1 , ·}. Is the theory of groups universal? (no justifi-cation required)
3. Show that T is universal iff T∀ |= T .
Exercise 1’.7. Two theories T and T ′ are companions if every model of T isincluded (not necessarily elementarily!) in a model of T ′, and vice-versa.
1. Find an interesting companion of the theory of orderings. Find an inter-esting companion of the theory of rings.
2. Prove that T and T∀ are companions.
3. Deduce that T and T ′ are companions if and only if T∀ and T ′∀ are com-panions.
4. Prove that T is universal if and only if every substructure of a model ofT is a model of T .
5. Prove that T∀ ⊆ T ′∀ if and only if every model of T ′ is included (notnecessarily elementarily, well-understood) in a model of T .
Exercise 1’.8.
1. SupposeM1 ≡M2. Show that there isM∗ such thatM1 andM2 embedintoM∗ elementarily.
2. Let N be a model which embeds elementarily into Mi for each i ∈ I.Show that there is a common elementary extension M∗ � Mi in whichall copies of N are glued togehter.
Exercise 1’.9. Let T be a theory and ϕ(x) be a formula. Prove that ϕ isequivalent modulo T to an existential formula if and only if:
for all modelsM and N of T withM⊆ N and a ∈M ,one hasM |= ϕ(a)⇒ N |= ϕ(a).
[Hint: one implication is trivial. For the other, study the set Ψ := {ψ(c) anexistential formula such that T |= ψ(c)→ ϕ(c)}. Prove that T1 := T ∪ {¬ψ(c) :ψ ∈ Ψ} ∪ {ϕ(c)} is not consistent.]
Exercise 1’.10. Show the converse of Theorem 1’.5.19, that is show: if T elim-inates quantifiers and M,N are ω-saturated models of T , then local isomor-phisms form an ∞-isomorphism family.
Exercise 1’.11. The language consists of a single binary relation symbol −.The theory of graphs is given by the axioms:
• ∀x ¬(x− x)
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• ∀x∀y (x− y → y − x).
A graph G is connected if for any x, y ∈ G, there is a path from x to y. (Thisdefinition does not sound first-order: we shall prove it.) Show that there is notheory whose models are exactly all connected graphs.
Exercise 1’.12. Consider the two groups Q and Q ⊕ Q. How isomorphic arethey (as groups)? Same question when they are considered as rings.
Exercise 1’.13. We consider the following <-structures:
A = . . .∐<
Z∐<
. . .︸ ︷︷ ︸Z
and B = A∐<
B
where X∐< Y means that the disjoint union is ordered in such a way that every
y ∈ Y lies above every x ∈ X.How isomorphic are A and B?
Exercise 1’.14. Let T be the {R}-theory asserting that R is an equivalencerelation. For any modelM, define the following numbers:
for each k ∈ N ∪ {∞}, mk ∈ N ∪ {∞} is the number of R-classeswith exactly k elements.
LetM and N be models with numbers (mk) and (nk) respectively. Determinehow isomorphicM and N are.
Exercise 1’.15. Let T be a complete theory with infinite models, and ϕ(x) bea formula. Suppose that there is a cardinal κ such that for any modelM, thereare at most κ elements of M satisfying ϕ(x). Show that there is a number nsuch that in any model, ϕ(x) has exactly n elements.
Exercise 1’.16. LetM be an L-structure. Prove that a definable subset thatis invariant under every automorphism of M is definable without parameters.[Hint: let X = ϕ(x, a) be the definable set and consider Ψ := {ψ(x) : M |=∀x ψ(x)→ ϕ(x, a)}. Form an La(x)-theory and prove it inconsistent...]
Exercise 1’.17. Recall that a theory is said to be universal if it has an axiom-atization by universal sentences. Also, if T is a theory, let T∀ be the set of allits consequences that are universal.
Two theories T and T ′ are companions if every model of T is included (notnecessarily elementarily!) in a model of T ′, and vice-versa.
1. Find interesting examples.
2. Prove that T and T∀ are companions.
3. Deduce that T and T ′ are companions if and only if T∀ and T ′∀ are com-panions.
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4. Prove that T is universal if and only if every sub-structure of a model ofT is a model of T .
5. Prove that T∀ ⊆ T ′∀ if and only if every model of T ′ is included (notnecessarily elementarily, well-understood) in a model of T .
Exercise 1’.18. Show that a non-principal ultraproduct is ω-saturated (it iseven ω1-saturated).
Exercise 1’.19. Let L = {<} and T be the following theory:
• ∀x ¬(x < x)
• ∀x∀y (x < y ∨ x = y ∨ y < x)
• ∀x∀y∀z (x < y ∧ y < z) → x < z
• ∀x∃y∀z (z > x)→ (z > y ∨ z = y)
• ∀x∃y∃z (y < x ∧ x < z)
1. Explain what the theory says (it has a fairly standard name).
2. Describe countable models of T (no justification required).
3. Describe ω-saturated models of T (no justification required).
4. Prove that any two ω-saturated models of T are ∞-isomorphic.
5. Show that it is not true that local isomorphisms between ω-saturatedmodels of T are ∞-isomorphisms.
6. Show that T is complete.
7. Does T eliminate quantifiers?
8. Is T ℵ0-categorical? Is it ℵ1-categorical?
9. Show that any model of T embeds into a model of DLO.
10. Show that any model of DLO embeds into a model of T .
Exercise 1’.20. Let L = {B,R,<} where B and R (“blue” and “red”) are unarypredicate symbols, and < is a binary relation symbol. Consider the followingconditions:
• every point is either blue or red, but not both
• < is an order relation
• above and below any blue point, there is a red point
• above and below any red point, there is a blue point
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• between any two points, there is a point
• between any two blue points, there is a red point
• between any two red points, there is a blue point
1. Write down L-sentences expressing the above axioms. Let T be the re-sulting L-theory.
2. Show that an order-preserving function need not be a local isomorphism.
3. Prove that any two models of T are ∞-isomorphic.
4. Show that T is complete.
5. Does T eliminate quantifiers?
6. Is T ℵ0-categorical? Is it ℵ1-categorical?
7. What happens if we drop the clause that between any two points there isa point?
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Index of Notions
Algebraically Closed Field*, 122ACFq, 122
AssignmentTruth Assignment, 11of the Variables, 51
Atomic Formula, see FormulaAxiom
Axioms of Equality, 57Deduction Rule, 16
Axiomatization of a Theory, 32
Back-and-ForthFinitary*, 108of Height ω*, 109Infinite, 107
Categoricity, 99Absolute Categoricity, 87
Compactnessin Topology*, 34
Complete Theory, 29, 71V-Complete, 67
ConsequenceSemantic Consequence, 12, 52Syntactic Consequence, 16, 57
Consistency, 17, 58
Decidable, 32Semi-Decidable, 32
Deduction, 16, 57D. Rulesfor Connectives, 16for ∀ and ∃, 57
Dense Linear Ordering, 104
ElementaryEquivalence, 92Inclusion, see InclusionMorphism, see MorphismSubstructure, see Substructure
Elimination Set, 111Equality Axioms, see Axiom
Filter*, 74
Finite Intersection Property*, 34Formula
Atomic Formula, 48Existential Formula, 90Quantifier-Free Formula, 90Universal Formula, 90Turning one into a Sentence, 91
Inclusion, 93Elementary Inclusion, 95
Interpretation, 51Isomorphism, 97
0-Isomorphism, 106n-Isomorphism*, 108ω-Isomorphism*, 109∞-Isomorphism, 107
Karp Family, 107Kripke Model*, 35
Languageof Propositional Logic, 8of Modal Logic*, 35First-Order Language, 46Second-Order Language, 85
Local Isomorphism, see Isom.
Model, 52, 91in Modal Logic*, see Kripke
Modus Ponens, 17Modus Tollens, 17Morphism, 97
Elementary Morphism, 98
Non-Standard Reals*, 73
Peano Arithmetic, 87
Quantification Rank, 91Quantifier Elimination, 111
Satisfactionin Propositional Logic, 12in Modal Logic*, 36
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in First-Order Logic, 51Satisfiability
in Propositional Logic, 12in First-Order Logic, 52
ω-Saturation, 116Sentence, 49Signature, 46Skolem Function, 101Standard Part*, 74Structure, 50Substitutable Term, see TermSubstructure, 93
Elementary Substructure, 95
Term, 47Substitutable Term, 53
Theoryin Propositional Logic, 9in First-Order Logic, 49of a First-Order Structure, 92
Truth Assignment, see AssignmentType, 115
of a Tuple, 109
Ultrafilter*, 75Ultraproduct*, 76
Variable, Free or Bound, 49
Weakening, 16Well-Formed Formula
in Propositional Logic, 8in First-Order Logic, 48
Witness, 63
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Index of Results
CategoricityAbsolute C. of PA, 87Morley’s C. Theorem*, 99ℵ1-C. of ACF*, 122ℵ0-C. of DLO, 104, 118ℵ1-C. of Th(N, 0, s), 119
Compactnessof Propositional Logic, 31of Modal Logic*, 40of First-Order Logic, 71fails in Higher Logic, 87
Completenessof Propositional Logic, 29of Modal Logic*, 38–40of First-Order Logic, 62
Completeness of a TheoryCriteria, 92, 103, 117of DLO, 105, 118of the Ordering on N, 120of the Successor Function, 119of ACFq*, 122
Decidability*Semi-D. of a Theory, 33, 70of a Complete Theory, 33, 71
Elimination Theorem, 111Excluded Middle, 19
Generalisation Theorem, 64
Łoś’ Theorem*, 77Łoś-Vaught Completeness Test, 103Löwenheim-Skolem Theorem
Descending Version, 102Ascending Version, 103Final Version, 103
ω-Saturated Models Do Exist, 116
QECriteria, 113, 117of DLO, 118of Th(N, 0, <), 120
of Th(N, 0, s), 119of ACF*, 122
Renaming Algorithm, 54
Skolem’s Paradox, 102Soundness
of Propositional Logic, 27of Modal Logic*, 37of First-Order Logic, 61
Tarski’s Test for �, 96
Unique Readability, 9, 48
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List of Lectures
Lecture 1 (Language and Wff’s; Unique Readability) . . . . . . . . . . . . 7Lecture 2 (The Semantics of Propositional Logic) . . . . . . . . . . . . . . 11Lecture 3 (Natural Deduction in Classical Logic) . . . . . . . . . . . . . . 15Lecture 4 (Reduction to Two Connectives (1/2)) . . . . . . . . . . . . . . 20Lecture 5 (Reduction to Two Connectives (2/2); Soundness) . . . . . . . . 24Lecture 6 (Completeness of Propositional Logic) . . . . . . . . . . . . . . 28Lecture 7 (Compactness of Propositional Logic) . . . . . . . . . . . . . . . 30Lecture 8 (Modal Logic*) . . . . . . . . . . . . . . . . . . . . . . . . . . . 35Lecture 9 (ME1) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40Lecture 10 (First-Order Languages; Terms and Formulas) . . . . . . . . . 45Lecture 11 (The Semantics of First-Order Logic) . . . . . . . . . . . . . . 50Lecture 12 (Substitutions) . . . . . . . . . . . . . . . . . . . . . . . . . . . 52Lecture 13 (Deductions; Simplifying the Language) . . . . . . . . . . . . . 57Lecture 14 (Soundness; Completeness (1/2)) . . . . . . . . . . . . . . . . . 61Lecture 15 (Completeness (2/2)) . . . . . . . . . . . . . . . . . . . . . . . 66Lecture 16 (Consequences; Compactness; Non-Standard Analysis) . . . . 70Lecture 17 (Proof of Compactness by Ultraproducts*) . . . . . . . . . . . 74Lecture 18 (Second-Order Logic) . . . . . . . . . . . . . . . . . . . . . . . 85Lecture 19 (ME2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88Lecture 20 (Models and Theories; Elementary Equivalence) . . . . . . . . 89Lecture 21 (Elementary Inclusion; Morphisms) . . . . . . . . . . . . . . . 94Lecture 22 (Löwenheim-Skolem Theorems) . . . . . . . . . . . . . . . . . 99Lecture 23 (Back-and-Forth Methods) . . . . . . . . . . . . . . . . . . . . 103Lecture 24 (Elimination Theorem) . . . . . . . . . . . . . . . . . . . . . . 111Lecture 25 (ω-Saturated Models) . . . . . . . . . . . . . . . . . . . . . . . 114Lecture 26 (Examples) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118
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