Mathematical Logic (MTH 302) : Quiz 2 Solutions

Problem 1. (a) Atoms in the Boolean algebra formed by F/ ≡ are (i) [p1 ∧ p2], (ii) [¬p1 ∧ p2],(iii) [p1 ∧ ¬p2] and (iv) [¬p1 ∧ ¬p2], where [·] denotes an equivalence class of wff. 4 marks

(b)F : {T, F}PV −→ {T, F}F/≡

v 7→ F (v)

where F (v) : F/ ≡ −→ {T, F} such that F (v)([α]) := v(α).

(i) For F (v) to be well defined, we must have [α] = [β]⇒ F (v)([α]) = F (v)([β]).

[α] = [β]

⇒ v(α) = v(β)

⇒ F (v)([α]) = F (v)([β])

Note that, we have to show the well definedness of F (v) and not the well definedness of F .2 marks

(ii) To show surjectivity, we must show that given a homomorphism h : F/ ≡ −→ {T, F},there exists a valuation v such that F (v) = h.Define v : PV −→ {T, F} such that v(p) := h([p]), for any p ∈ PV . Extend v over wffs.Claim: v(α) = h([α]), for any wff α. Use induction on the number of connectives in α.The basis case is obtained by definition of v. Suppose v(α) = h([α]) where α has n orless connectives. If α has n + 1 connectives, α is of the form ¬β or β ∧ γ or β ∨ γ orα→ β, where β and γ have n or less connectives. Let α = ¬β. We have v(β) = h([β]) (byinduction hypothesis as β has n or less connectives.

v(α) = v(¬β) = ¬v(β) = ¬h([β]) = h(¬[β]) = h([¬β]) = h([α]).

Similarly, for ∨, ∧ and → cases.Thus, F (v)([α]) = v(α) (given)= h([α]) (by claim).Thus F (v) = h. 4 marks

Problem 2. (i) (a) No, ∀x is followed here by a function symbol f , and no wff begins with afunction symbol. 1 mark

(b) Yes, as ∀xqxfxxfxxxx is of the form ∀x q(t1, t2) where t1 = f(x, x, f(x, x, x)) and t2 = x.1 mark

(ii) (a) No. On replacing y by fxfxxxx in (∃x qxy), the occurrence of x in fxfxxxx will becomebound as x is bound in (∃x qxy). 1 mark

(b) Yes. On replacing y by fxfxxxx in ∃z (qyz → pfxfxxxx), no occurrence of a variable offxfxxxx becomes bound as x is free in ∃z (qyz → pfxfxxxx). 1 mark

1

Problem 3. (a) Consider an interpretation I and a valuation v in FOL.To prove that I, v � (∀x α∨∀x β)→ ∀x(α∨β) i.e. if I, v � (∀x α∨∀x β) then I, v � ∀x (α∨β).Let I, v � (∀x α ∨ ∀x β) i.e. I, v � ∀x α or I, v � ∀x β. Claim: I, v � ∀x (α ∨ β)If I, v � ∀x α⇒ for all v′, x−equivalent to v, I, v′ � α⇒ for all v′, x−equivalent to v, I, v′ � (α ∨ β)⇒ I, v � ∀x (α ∨ β)Similarly, for I, v � ∀x β. 3 marks

(b) Consider the FOL with variable x and two unary predicate symbols p and q. Define the domainof interpretation I as N, the set of natural numbers and sets R1 as set of even numbers and R2

as set of odd numbers such that p is assigned R1 and q is assigned R2 in the interpretation I.Consider the wff ∀x (px∨ qx)→ (∀x px∨ ∀x qx). Take any arbitrary valuation v i.e. v(x) ∈ N.For all valuation v′, x-equivalent to v, we have I, v′ � (px ∨ qx) ⇔ v′(x) ∈ R1 or v′(x) ∈ R2,that is always true as v′(x) ∈ N is either even or odd. Thus I, v � ∀x (px ∨ qx).However, I, v � (∀x px ∨ ∀x qx) ⇔ I, v � ∀x px or I, v � ∀x qx.Let us assume I, v � ∀x px.Then For all valuations v′′, x-equivalent to v, I, v′′ � px i.e. v′′(x) ∈ R1 i.e. v′′(x) is even,which is not true always as we can define v′′ such that v′′(x) is odd. So, I, v 2 ∀x px.Similarly, I, v 2 ∀x qx. So I, v 2 (∀x px ∨ ∀x qx).Thus I, v 2 ∀x (px ∨ qx)→ (∀x px ∨ ∀x qx). 3 marks

2