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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

New Rules

Assumptions Line Formula Justification

1 (1) (∀x)Fxx A

1 (2) Fbb 1 ∀E

New Rules

Assumptions Line Formula Justification

1 (1) (∀x)Fxx A

1 (2) Fbb 1 ∀E

1 (3) (∀x)Fxb 2 ∀I

New Rules

Assumptions Line Formula Justification

1 (1) (∀x)Fxx A

1 (2) Fbb 1 ∀E

1 (3) (∀x)Fxb 2 ∀I

NO!

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!

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

New Rules

Assumptions Line Formula Justification

1 (1) Fa A

1 (2) (∀x)Fx 1 ∀I

New Rules

Assumptions Line Formula Justification

1 (1) Fa A

1 (2) (∀x)Fx 1 ∀I

NO!

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!

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

New Rules

Assumptions Line Formula Justification

1 (1) (∀x)(Fa → Gx) A

2 (2) Fa A (for CP)

New Rules

Assumptions Line Formula Justification

1 (1) (∀x)(Fa → Gx) A

2 (2) Fa A (for CP)

1 (3) (Fa → Gb) 1 ∀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

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

New Rules

Assumptions Line Formula Justification

1 (1) Fa A

New Rules

Assumptions Line Formula Justification

1 (1) Fa A

1 (2) (∃x)Fx 1 ∃I

New Rules

Let’s try an example:

{ } (Fbb → ((∃x)Fxb ᴧ (∃x)Fbx)

New Rules

Assumptions Line Formula Justification

1 (1) Fbb A (for CP)

New Rules

Assumptions Line Formula Justification

1 (1) Fbb A (for CP)

1 (2) (∃x)Fxb 1 ∃I

New Rules

Assumptions Line Formula Justification

1 (1) Fbb A (for CP)

1 (2) (∃x)Fxb 1 ∃I

1 (3) (∃x)Fbx 1 ∃I

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

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

New Rules

Assumptions Line Formula Justification

1 (1) (∀x)(∀y)Rxy A

1 (2) (∀y)Ray 1 ∀E

New Rules

Assumptions Line Formula Justification

1 (1) (∀x)(∀y)Rxy A

1 (2) (∀y)Ray 1 ∀E

1 (3) Rab 2 ∀E

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

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

New Rules

Assumptions Line Formula Justification

1 (1) (∃x)Fx A

2 (2) Fa A (for CP)

New Rules

Assumptions Line Formula Justification

1 (1) (∃x)Fx A

2 (2) Fa A (for CP)

(3) (Fa → Fa) 2 CP

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

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

New Rules

Assumptions Line Formula Justification

1 (1) (∃x)Gx A

2 (2) ~Ga A

2 (3) (Ga → Gb) 2 Lemma

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

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)

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

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

New Rules

Assumptions Line Formula Justification

1 (1) (∃x)~Fx A

New Rules

Assumptions Line Formula Justification

1 (1) (∃x)~Fx A

2 (2) (∀x)Fx A*

New Rules

Assumptions Line Formula Justification

1 (1) (∃x)~Fx A

2 (2) (∀x)Fx A*

3 (3) ~Fa A (for 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

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

New Rules

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.