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First-Order Logic
Natural Deduction, Part 2
Review
Let ϕ be a formula, let x be an arbitrary variable, and let c be an arbitrary constant.
ϕ[x/c] denotes the result of replacing all free occurrences of x in ϕ with c.
Review
(Gzz → Hxz)
(∀x)(Sx → Px)
(Fy ᴧ (∃y)(By ᴧ Dx))
(Rxx ᴧ (∃x)Fx)
(~Dy → Kz)
Review
(Gzz → Hxz)[z/d]
(∀x)(Sx → Px)
(Fy ᴧ (∃y)(By ᴧ Dx))
(Rxx ᴧ (∃x)Fx)
(~Dy → Kz)
Review
(Gzz → Hxz)[z/d]
(∀x)(Sx → Px)
(Fy ᴧ (∃y)(By ᴧ Dx))
(Rxx ᴧ (∃x)Fx)
(~Dy → Kz)
(Gdd → Hxd)
Review
(Gzz → Hxz)[z/d]
(∀x)(Sx → Px)[y/e]
(Fy ᴧ (∃y)(By ᴧ Dx))
(Rxx ᴧ (∃x)Fx)
(~Dy → Kz)
(Gdd → Hxd)
Review
(Gzz → Hxz)[z/d]
(∀x)(Sx → Px)[y/e]
(Fy ᴧ (∃y)(By ᴧ Dx))
(Rxx ᴧ (∃x)Fx)
(~Dy → Kz)
(Gdd → Hxd)
(∀x)(Sx → Px)
Review
(Gzz → Hxz)[z/d]
(∀x)(Sx → Px)[y/e]
(Fy ᴧ (∃y)(By ᴧ Dx))[y/c]
(Rxx ᴧ (∃x)Fx)
(~Dy → Kz)
(Gdd → Hxd)
(∀x)(Sx → Px)
Review
(Gzz → Hxz)[z/d]
(∀x)(Sx → Px)[y/e]
(Fy ᴧ (∃y)(By ᴧ Dx))[y/c]
(Rxx ᴧ (∃x)Fx)
(~Dy → Kz)
(Gdd → Hxd)
(∀x)(Sx → Px)
(Fc ᴧ (∃y)(By ᴧ Dx))
Review
(Gzz → Hxz)[z/d]
(∀x)(Sx → Px)[y/e]
(Fy ᴧ (∃y)(By ᴧ Dx))[y/c]
(Rxx ᴧ (∃x)Fx)[x/b]
(~Dy → Kz)
(Gdd → Hxd)
(∀x)(Sx → Px)
(Fc ᴧ (∃y)(By ᴧ Dx))
Review
(Gzz → Hxz)[z/d]
(∀x)(Sx → Px)[y/e]
(Fy ᴧ (∃y)(By ᴧ Dx))[y/c]
(Rxx ᴧ (∃x)Fx)[x/b]
(~Dy → Kz)
(Gdd → Hxd)
(∀x)(Sx → Px)
(Fc ᴧ (∃y)(By ᴧ Dx))
(Rbb ᴧ (∃x)Fx)
Review
(Gzz → Hxz)[z/d]
(∀x)(Sx → Px)[y/e]
(Fy ᴧ (∃y)(By ᴧ Dx))[y/c]
(Rxx ᴧ (∃x)Fx)[x/b]
(~Dy → Kz)[y/a]
(Gdd → Hxd)
(∀x)(Sx → Px)
(Fc ᴧ (∃y)(By ᴧ Dx))
(Rbb ᴧ (∃x)Fx)
Review
(Gzz → Hxz)[z/d]
(∀x)(Sx → Px)[y/e]
(Fy ᴧ (∃y)(By ᴧ Dx))[y/c]
(Rxx ᴧ (∃x)Fx)[x/b]
(~Dy → Kz)[y/a]
(Gdd → Hxd)
(∀x)(Sx → Px)
(Fc ᴧ (∃y)(By ᴧ Dx))
(Rbb ᴧ (∃x)Fx)
(~Da → Kz)
Review
Universal Elimination (∀E)
Suppose (∀x)ϕ is a well-formed formula for some variable x, and suppose c is an arbitrary constant.
ϕ[x/c]
(∀x)ϕ
Review
{ (∀x)(∀y)Rxy } Raa
Review
Assumptions Line Formula Justification
1 (1) (∀x)(∀y)Rxy A
Review
Assumptions Line Formula Justification
1 (1) (∀x)(∀y)Rxy A
1 (2) (∀y)Ray 1 ∀E
Review
Assumptions Line Formula Justification
1 (1) (∀x)(∀y)Rxy A
1 (2) (∀y)Ray 1 ∀E
1 (3) Raa 2 ∀E
New Rules
Universal Introduction (∀I)
Suppose ϕ is a well-formed formula, and c is a constant that does not appear in ϕ. Then
(∀x)ϕ
ϕ[x/c]
So long as ϕ[x/c] does not depend on any formula containing c.
New Rules
The rule for universal introduction has two clauses that constrain how we can introduce universal quantifiers.
The formula ϕ[x/c] cannot depend on any formula containing c.
The constant c cannot appear in ϕ.
New Rules
The easiest way to see how the constraints on universal introduction work is to see how they might be violated.
New Rules
The easiest way to see how the constraints on universal introduction work is to see how they might be violated. Let’s see a “proof” that violates the first constraint:
The constant c cannot appear in ϕ.
New Rules
Assumptions Line Formula Justification
1 (1) (∀x)Fxx A
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Assumptions Line Formula Justification
1 (1) (∀x)Fxx A
1 (2) Fbb 1 ∀E
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Assumptions Line Formula Justification
1 (1) (∀x)Fxx A
1 (2) Fbb 1 ∀E
1 (3) (∀x)Fxb 2 ∀I
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Assumptions Line Formula Justification
1 (1) (∀x)Fxx A
1 (2) Fbb 1 ∀E
1 (3) (∀x)Fxb 2 ∀I
NO!
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Assumptions Line Formula Justification
1 (1) (∀x)Fxx A
1 (2) Fbb 1 ∀E
1 (3) (∀x)Fxb 2 ∀I
NO!
The formula Fxb contains the constant b!
New Rules
Assumptions Line Formula Justification
1 (1) (∀x)Fxx A
1 (2) Fbb 1 ∀E
1 (3) (∀x)Fxb 2 ∀I
NO!
The formula Fxb contains the constant b!
Construct a small world showing invalidity.
New Rules
And now a “proof” that violates the second constraint:
The formula ϕ[x/c] cannot depend on any formula containing c.
New Rules
Assumptions Line Formula Justification
1 (1) Fa A
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Assumptions Line Formula Justification
1 (1) Fa A
1 (2) (∀x)Fx 1 ∀I
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Assumptions Line Formula Justification
1 (1) Fa A
1 (2) (∀x)Fx 1 ∀I
NO!
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Assumptions Line Formula Justification
1 (1) Fa A
1 (2) (∀x)Fx 1 ∀I
NO!
The formula Fa depends on a formula that contains the constant a, namely itself!
New Rules
Assumptions Line Formula Justification
1 (1) Fa A
1 (2) (∀x)Fx 1 ∀I
NO!
The formula Fa depends on a formula that contains the constant a, namely itself!
Construct a small world showing invalidity.
New Rules
Now, let’s see a correct example:
{ (∀x)(Fa → Gx) } (Fa → (∀x)Gx)
New Rules
Assumptions Line Formula Justification
1 (1) (∀x)(Fa → Gx) A
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Assumptions Line Formula Justification
1 (1) (∀x)(Fa → Gx) A
2 (2) Fa A (for CP)
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Assumptions Line Formula Justification
1 (1) (∀x)(Fa → Gx) A
2 (2) Fa A (for CP)
1 (3) (Fa → Gb) 1 ∀E
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Assumptions Line Formula Justification
1 (1) (∀x)(Fa → Gx) A
2 (2) Fa A (for CP)
1 (3) (Fa → Gb) 1 ∀E
1,2 (4) Gb 2,3 →E
New Rules
Assumptions Line Formula Justification
1 (1) (∀x)(Fa → Gx) A
2 (2) Fa A (for CP)
1 (3) (Fa → Gb) 1 ∀E
1,2 (4) Gb 2,3 →E
1,2 (5) (∀x)Gx 4 ∀I
New Rules
Assumptions Line Formula Justification
1 (1) (∀x)(Fa → Gx) A
2 (2) Fa A (for CP)
1 (3) (Fa → Gb) 1 ∀E
1,2 (4) Gb 2,3 →E
1,2 (5) (∀x)Gx 4 ∀I
1 (6) (Fa → (∀x)Gx) 2,5 CP
New Rules
Existential Introduction (∃I)
Suppose (∃x)ϕ is a well-formed formula for some variable x, and suppose c is an arbitrary constant.
(∃x)ϕ
ϕ[x/c]
New Rules
Let’s try an example:
{ Fa } (∃x)Fx
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Assumptions Line Formula Justification
1 (1) Fa A
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Assumptions Line Formula Justification
1 (1) Fa A
1 (2) (∃x)Fx 1 ∃I
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Let’s try an example:
{ } (Fbb → ((∃x)Fxb ᴧ (∃x)Fbx)
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Assumptions Line Formula Justification
1 (1) Fbb A (for CP)
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Assumptions Line Formula Justification
1 (1) Fbb A (for CP)
1 (2) (∃x)Fxb 1 ∃I
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Assumptions Line Formula Justification
1 (1) Fbb A (for CP)
1 (2) (∃x)Fxb 1 ∃I
1 (3) (∃x)Fbx 1 ∃I
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Assumptions Line Formula Justification
1 (1) Fbb A (for CP)
1 (2) (∃x)Fxb 1 ∃I
1 (3) (∃x)Fbx 1 ∃I
1 (4) ((∃x)Fxb ᴧ (∃x)Fbx) 2,3 ᴧI
New Rules
Assumptions Line Formula Justification
1 (1) Fbb A (for CP)
1 (2) (∃x)Fxb 1 ∃I
1 (3) (∃x)Fbx 1 ∃I
1 (4) ((∃x)Fxb ᴧ (∃x)Fbx) 2,3 ᴧI
(5) (1) → (4) 1,4 CP
New Rules
And one more:
{ (∀x)(∀y)Rxy } (∃x)(∃y)Rxy)
New Rules
Assumptions Line Formula Justification
1 (1) (∀x)(∀y)Rxy A
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Assumptions Line Formula Justification
1 (1) (∀x)(∀y)Rxy A
1 (2) (∀y)Ray 1 ∀E
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Assumptions Line Formula Justification
1 (1) (∀x)(∀y)Rxy A
1 (2) (∀y)Ray 1 ∀E
1 (3) Rab 2 ∀E
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Assumptions Line Formula Justification
1 (1) (∀x)(∀y)Rxy A
1 (2) (∀y)Ray 1 ∀E
1 (3) Rab 2 ∀E
1 (4) (∃y)Ray 3 ∃I
New Rules
Assumptions Line Formula Justification
1 (1) (∀x)(∀y)Rxy A
1 (2) (∀y)Ray 1 ∀E
1 (3) Rab 2 ∀E
1 (4) (∃y)Ray 3 ∃I
1 (5) (∃x)(∃y)Rxy 4 ∃I
Putting It Together
Give a proof of the conclusion of this argument from its premisses:
(∀x)(∃y)Rxy
(∃x)(∃y)Rxy → (∀x)(Fx → Gx)
((∀x)Fx → (∀x)Gx)
Putting It TogetherAssumptions Line Formula Justification
1 (1) (∀x)(∃y)Rxy A
2 (2) (∃x)(∃y)Rxy → (∀x)(Fx → Gx) A
Putting It TogetherAssumptions Line Formula Justification
1 (1) (∀x)(∃y)Rxy A
2 (2) (∃x)(∃y)Rxy → (∀x)(Fx → Gx) A
3 (3) (∀x)Fx ACP
Putting It TogetherAssumptions Line Formula Justification
1 (1) (∀x)(∃y)Rxy A
2 (2) (∃x)(∃y)Rxy → (∀x)(Fx → Gx) A
3 (3) (∀x)Fx ACP
1 (4) (∃y)Ray 1 ∀E
Putting It TogetherAssumptions Line Formula Justification
1 (1) (∀x)(∃y)Rxy A
2 (2) (∃x)(∃y)Rxy → (∀x)(Fx → Gx) A
3 (3) (∀x)Fx ACP
1 (4) (∃y)Ray 1 ∀E
1 (5) (∃x)(∃y)Rxy 4 ∃I
Putting It TogetherAssumptions Line Formula Justification
1 (1) (∀x)(∃y)Rxy A
2 (2) (∃x)(∃y)Rxy → (∀x)(Fx → Gx) A
3 (3) (∀x)Fx ACP
1 (4) (∃y)Ray 1 ∀E
1 (5) (∃x)(∃y)Rxy 4 ∃I
1,2 (6) (∀x)(Fx → Gx) 2,5 →E
Putting It TogetherAssumptions Line Formula Justification
1 (1) (∀x)(∃y)Rxy A
2 (2) (∃x)(∃y)Rxy → (∀x)(Fx → Gx) A
3 (3) (∀x)Fx ACP
1 (4) (∃y)Ray 1 ∀E
1 (5) (∃x)(∃y)Rxy 4 ∃I
1,2 (6) (∀x)(Fx → Gx) 2,5 →E
3 (7) Fb 3 ∀E
Putting It TogetherAssumptions Line Formula Justification
1 (1) (∀x)(∃y)Rxy A
2 (2) (∃x)(∃y)Rxy → (∀x)(Fx → Gx) A
3 (3) (∀x)Fx ACP
1 (4) (∃y)Ray 1 ∀E
1 (5) (∃x)(∃y)Rxy 4 ∃I
1,2 (6) (∀x)(Fx → Gx) 2,5 →E
3 (7) Fb 3 ∀E
1,2 (8) (Fb → Gb) 6 ∀E
Putting It TogetherAssumptions Line Formula Justification
1 (1) (∀x)(∃y)Rxy A
2 (2) (∃x)(∃y)Rxy → (∀x)(Fx → Gx) A
3 (3) (∀x)Fx ACP
1 (4) (∃y)Ray 1 ∀E
1 (5) (∃x)(∃y)Rxy 4 ∃I
1,2 (6) (∀x)(Fx → Gx) 2,5 →E
3 (7) Fb 3 ∀E
1,2 (8) (Fb → Gb) 6 ∀E
1,2,3 (9) Gb 7,8 →E
Putting It TogetherAssumptions Line Formula Justification
1 (1) (∀x)(∃y)Rxy A
2 (2) (∃x)(∃y)Rxy → (∀x)(Fx → Gx) A
3 (3) (∀x)Fx ACP
1 (4) (∃y)Ray 1 ∀E
1 (5) (∃x)(∃y)Rxy 4 ∃I
1,2 (6) (∀x)(Fx → Gx) 2,5 →E
3 (7) Fb 3 ∀E
1,2 (8) (Fb → Gb) 6 ∀E
1,2,3 (9) Gb 7,8 →E
1,2,3 (10) (∀x)Gx 9 ∀I
Putting It TogetherAssumptions Line Formula Justification
1 (1) (∀x)(∃y)Rxy A
2 (2) (∃x)(∃y)Rxy → (∀x)(Fx → Gx) A
3 (3) (∀x)Fx ACP
1 (4) (∃y)Ray 1 ∀E
1 (5) (∃x)(∃y)Rxy 4 ∃I
1,2 (6) (∀x)(Fx → Gx) 2,5 →E
3 (7) Fb 3 ∀E
1,2 (8) (Fb → Gb) 6 ∀E
1,2,3 (9) Gb 7,8 →E
1,2,3 (10) (∀x)Gx 9 ∀I
1,2 (11) ((∀x)Fx → (∀x)Gx) 3,10 CP
Next Time
We’ll look at one more rule for the existential quantifier and two rules for the identity relation.
New Rules
Existential Elimination (∃E)
Suppose φ and ψ are well-formed formulas in which the constant c does not appear. Then
ψ
(φ[x/c] → ψ)
So long as (φ[x/c] → ψ) does not depend onany formula containing c.
(∃x)φ
New Rules
As with Universal Introduction, Existential Elimination has two constraints on its application:
The formula (φ[x/c] → ψ) cannot rest on any formula containing c.
The constant c cannot appear in φ or ψ.
New Rules
Again, it will be easier to see how the constraints work by seeing how they might be violated.
New Rules
Consider a case where the first constraint is violated:
The constant c cannot appear in φ or ψ.
New Rules
Assumptions Line Formula Justification
1 (1) (∃x)Fx A
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Assumptions Line Formula Justification
1 (1) (∃x)Fx A
2 (2) Fa A (for CP)
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Assumptions Line Formula Justification
1 (1) (∃x)Fx A
2 (2) Fa A (for CP)
(3) (Fa → Fa) 2 CP
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Assumptions Line Formula Justification
1 (1) (∃x)Fx A
2 (2) Fa A (for CP)
(3) (Fa → Fa) 2 CP
1 (4) Fa 1,3 ∃E
New Rules
Assumptions Line Formula Justification
1 (1) (∃x)Fx A
2 (2) Fa A (for CP)
(3) (Fa → Fa) 2 CP
1 (4) Fa 1,3 ∃E
NO!
New Rules
Assumptions Line Formula Justification
1 (1) (∃x)Fx A
2 (2) Fa A (for CP)
(3) (Fa → Fa) 2 CP
1 (4) Fa 1,3 ∃E
NO!
Here, the constant a in Fx[x/a] on line (2) also appears in ψ = Fa.
New Rules
Now, consider a case where the second constraint is violated:
The formula (φ[x/c] → ψ) cannot rest on any formula containing c.
New Rules
Assumptions Line Formula Justification
1 (1) (∃x)Gx A
2 (2) ~Ga A
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Assumptions Line Formula Justification
1 (1) (∃x)Gx A
2 (2) ~Ga A
2 (3) (Ga → Gb) 2 Lemma
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Assumptions Line Formula Justification
1 (1) (∃x)Gx A
2 (2) ~Ga A
2 (3) (Ga → Gb) 2 Lemma
1,2 (4) Gb 1,4 ∃E
New Rules
Assumptions Line Formula Justification
1 (1) (∃x)Gx A
2 (2) ~Ga A
2 (3) (Ga → Gb) 2 Lemma
1,2 (4) Gb 1,4 ∃E
NO!
New Rules
Assumptions Line Formula Justification
1 (1) (∃x)Gx A
2 (2) ~Ga A
2 (3) (Ga → Gb) 2 Lemma
1,2 (4) Gb 1,4 ∃E
NO!
The formula (Gx[x/a] → ψ) on line (3) depends on a formula that contains the constant a.
New Rules
Now, let’s see a correct example:
{ (∀x)(Fx → Gb) } ((∃x)Fx → Gb)
New Rules
Assumptions Line Formula Justification
1 (1) (∀x)(Fx → Gb) A
New Rules
Assumptions Line Formula Justification
1 (1) (∀x)(Fx → Gb) A
2 (2) (∃x)Fx A (for CP)
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Assumptions Line Formula Justification
1 (1) (∀x)(Fx → Gb) A
2 (2) (∃x)Fx A (for CP)
1 (3) (Fa → Gb) 1 ∀E
New Rules
Assumptions Line Formula Justification
1 (1) (∀x)(Fx → Gb) A
2 (2) (∃x)Fx A (for CP)
1 (3) (Fa → Gb) 1 ∀E
1,2 (4) Gb 2,3 ∃E
New Rules
Assumptions Line Formula Justification
1 (1) (∀x)(Fx → Gb) A
2 (2) (∃x)Fx A (for CP)
1 (3) (Fa → Gb) 1 ∀E
1,2 (4) Gb 2,3 ∃E
1 (5) ((∃x)Fx → Gb) 2,4 CP
New Rules
Let’s try a more complicated example:
{ (∃x)~Fx } ~(∀x)Fx
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Assumptions Line Formula Justification
1 (1) (∃x)~Fx A
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Assumptions Line Formula Justification
1 (1) (∃x)~Fx A
2 (2) (∀x)Fx A*
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Assumptions Line Formula Justification
1 (1) (∃x)~Fx A
2 (2) (∀x)Fx A*
3 (3) ~Fa A (for CP)
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Assumptions Line Formula Justification
1 (1) (∃x)~Fx A
2 (2) (∀x)Fx A*
3 (3) ~Fa A (for CP)
2 (4) Fa 2 ∀E
New Rules
Assumptions Line Formula Justification
1 (1) (∃x)~Fx A
2 (2) (∀x)Fx A*
3 (3) ~Fa A (for CP)
2 (4) Fa 2 ∀E
(5) ((∀x)Fx → Fa) 2,4 CP
New Rules
Assumptions Line Formula Justification
1 (1) (∃x)~Fx A
2 (2) (∀x)Fx A*
3 (3) ~Fa A (for CP)
2 (4) Fa 2 ∀E
(5) ((∀x)Fx → Fa) 2,4 CP
3 (6) ((∀x)Fx → ~Fa) 3 →I
New Rules
Assumptions Line Formula Justification
1 (1) (∃x)~Fx A
2 (2) (∀x)Fx A*
3 (3) ~Fa A (for CP)
2 (4) Fa 2 ∀E
(5) ((∀x)Fx → Fa) 2,4 CP
3 (6) ((∀x)Fx → ~Fa) 3 →I
3 (7) ~(∀x)Fx 5,6 ~I
New Rules
Assumptions Line Formula Justification
1 (1) (∃x)~Fx A
2 (2) (∀x)Fx A*
3 (3) ~Fa A (for CP)
2 (4) Fa 2 ∀E
(5) ((∀x)Fx → Fa) 2,4 CP
3 (6) ((∀x)Fx → ~Fa) 3 →I
3 (7) ~(∀x)Fx 5,6 ~I
(8) (~Fa → ~(∀x)Fx) 3,7 CP
New Rules
Assumptions Line Formula Justification
1 (1) (∃x)~Fx A
2 (2) (∀x)Fx A*
3 (3) ~Fa A (for CP)
2 (4) Fa 2 ∀E
(5) ((∀x)Fx → Fa) 2,4 CP
3 (6) ((∀x)Fx → ~Fa) 3 →I
3 (7) ~(∀x)Fx 5,6 ~I
(8) (~Fa → ~(∀x)Fx) 3,7 CP
1 (9) ~(∀x)Fx 1,8 ∃E
New Rules
Identity is a very special relation. And we give it special treatment.
Unlike the relations “… is taller than ---,” “… is made of ---,” and so on, the identity relation gets its own introduction and elimination rules.
New Rules
Identity Introduction (=I)
On any line in a proof, you may write the formula (∀x)(x = x) without writing anything in the assumption column.
(∀x)(x = x)
New Rules
Let’s see an example:
{ } ((a=a) ᴧ (b=b))
New Rules
Assumptions Line Formula Justification
(1) (∀x)(x = x) =I
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Assumptions Line Formula Justification
(1) (∀x)(x = x) =I
(2) (a = a) 1 ∀E
New Rules
Assumptions Line Formula Justification
(1) (∀x)(x = x) =I
(2) (a = a) 1 ∀E
(3) (b = b) 1 ∀E
New Rules
Assumptions Line Formula Justification
(1) (∀x)(x = x) =I
(2) (a = a) 1 ∀E
(3) (b = b) 1 ∀E
(4) ((a = a) ᴧ (b = b)) 2,3 ᴧI
New Rules
The identity introduction rule is simple. In order to state the identity elimination rule, we need a new kind of operation: partial substitution.
New Rules
Partial Substitution:
Let b and c be arbitrary constants, and let ϕbe an arbitrary formula. Then ϕ[[b/c]] represents any formula obtained from ϕ by replacing some occurrences of the constant b with the constant c.
New Rules
Examples of Partial Substitution:
Ga
Haa
Hab
(a=b)
(∀x)Hxb
New Rules
Examples of Partial Substitution:
Ga[[a/b]]
Haa
Hab
(a=b)
(∀x)Hxb
New Rules
Examples of Partial Substitution:
Ga[[a/b]]
Haa
Hab
(a=b)
(∀x)Hxb
Gb
New Rules
Examples of Partial Substitution:
Ga[[a/b]]
Haa[[a/b]]
Hab
(a=b)
(∀x)Hxb
Gb
New Rules
Examples of Partial Substitution:
Ga[[a/b]]
Haa[[a/b]]
Hab
(a=b)
(∀x)Hxb
Gb
Hab, Hba, Hbb
New Rules
Examples of Partial Substitution:
Ga[[a/b]]
Haa[[a/b]]
Hab[[b/c]]
(a=b)
(∀x)Hxb
Gb
Hab, Hba, Hbb
New Rules
Examples of Partial Substitution:
Ga[[a/b]]
Haa[[a/b]]
Hab[[b/c]]
(a=b)
(∀x)Hxb
Gb
Hab, Hba, Hbb
Hac
New Rules
Examples of Partial Substitution:
Ga[[a/b]]
Haa[[a/b]]
Hab[[b/c]]
(a=b)[[b/c]]
(∀x)Hxb
Gb
Hab, Hba, Hbb
Hac
New Rules
Examples of Partial Substitution:
Ga[[a/b]]
Haa[[a/b]]
Hab[[b/c]]
(a=b)[[b/c]]
(∀x)Hxb
Gb
Hab, Hba, Hbb
Hac
(a=c)
New Rules
Examples of Partial Substitution:
Ga[[a/b]]
Haa[[a/b]]
Hab[[b/c]]
(a=b)[[b/c]]
(∀x)Hxb[[b/a]]
Gb
Hab, Hba, Hbb
Hac
(a=c)
New Rules
Examples of Partial Substitution:
Ga[[a/b]]
Haa[[a/b]]
Hab[[b/c]]
(a=b)[[b/c]]
(∀x)Hxb[[b/a]]
Gb
Hab, Hba, Hbb
Hac
(a=c)
(∀x)Hxa
New Rules
Identity Elimination (=E)
Let b and c be arbitrary constants, and let ϕbe an arbitrary formula. Then
ϕ[[b/c]]
(b = c)
ϕ
ϕ[[b/c]]
(c = b)
ϕ
New Rules
Let’s see an example:
{ Gab, (b = c) } Gac
New Rules
Assumptions Line Formula Justification
1 (1) Gab A
2 (2) (b = c) A
New Rules
Assumptions Line Formula Justification
1 (1) Gab A
2 (2) (b = c) A
1,2 (3) Gac 1,2 =E
New Rules
Let’s see another example:
{ (a = b) } (b = a)
New Rules
Assumptions Line Formula Justification
1 (1) (a = b) A
New Rules
Assumptions Line Formula Justification
1 (1) (a = b) A
(2) (∀x)(x = x) =I
New Rules
Assumptions Line Formula Justification
1 (1) (a = b) A
(2) (∀x)(x = x) =I
(3) (a = a) 2 ∀E
New Rules
Assumptions Line Formula Justification
1 (1) (a = b) A
(2) (∀x)(x = x) =I
(3) (a = a) 2 ∀E
1 (4) (b = a) 1,3 =E
New Rules
Okay, last example:
{ Ha, ~Hb } ~(a = b)
New Rules
Okay, last example:
{ Ha, ~Hb } (a ≠ b)
We’ll write (a ≠ b) instead of ~(a = b).
New Rules
Assumptions Line Formula Justification
1 (1) Ha A
2 (2) ~Hb A
New Rules
Assumptions Line Formula Justification
1 (1) Ha A
2 (2) ~Hb A
3 (3) (a = b) A*
New Rules
Assumptions Line Formula Justification
1 (1) Ha A
2 (2) ~Hb A
3 (3) (a = b) A*
2,3 (4) ~Ha 2,3 =E
New Rules
Assumptions Line Formula Justification
1 (1) Ha A
2 (2) ~Hb A
3 (3) (a = b) A*
2,3 (4) ~Ha 2,3 =E
2 (5) ((a = b) → ~Ha) 3,4 CP
New Rules
Assumptions Line Formula Justification
1 (1) Ha A
2 (2) ~Hb A
3 (3) (a = b) A*
2,3 (4) ~Ha 2,3 =E
2 (5) ((a = b) → ~Ha) 3,4 CP
1 (6) ((a = b) → Ha) 1 →I
New Rules
Assumptions Line Formula Justification
1 (1) Ha A
2 (2) ~Hb A
3 (3) (a = b) A*
2,3 (4) ~Ha 2,3 =E
2 (5) ((a = b) → ~Ha) 3,4 CP
1 (6) ((a = b) → Ha) 1 →I
1,2 (7) (a ≠ b) 5,6 ~I
Next Time
We’ll start thinking about set theory.