Functions Num : N N and Sub : N 3 N , Num( a ) := a , Sub( , - - PowerPoint PPT Presentation

Reminder of Num, Sub, ThmN Functions Num : N → N and Sub : N3 → N, Num(a) := a , Sub(ϕ, x, t) := ϕx

t ,

defined by ∆-formulas Num(x, y) and Sub(x1, x2, x3, y). The set ThmN ⊆ N, ThmN := {ϕ : N ⊢ ϕ}, is defined by a Σ-formula ThmN(x): ThmN(x) := ∃d DeductionN(d, x) where DeductionN(d, x) is a ∆-formula expressing “d = δ1, . . . , δn where (δ1, . . . , δn) is a deduction from N and x = δn”.

Representable ⇒ Σ-Definable Previously, we showed that every ∆-definable set is representable. (This is a straightforward corollary of Proposition 5.3.13: N proves every Σ- sentence which is true in N.) Next, we show that every representable set is Σ-definable.

Proposition 6.3.3. If A ⊆ Nk is representable, then A is Σ-definable.

• Proof. Let A ⊆ N for simplicity and assume that ϕ(v1) represents A.

(Without loss of generality, we take v1 to be the free variable of ϕ.)

Proposition 6.3.3. If A ⊆ Nk is representable, then A is Σ-definable.

• Proof. Let A ⊆ N for simplicity and assume that ϕ(v1) represents A.

Let β(x) be the following Σ-formula: β(x) :≡ ∃x∃y

• Num(x, y) ∧ Sub(ϕ, v1, y, z) ∧ ThmN(z)
• .

Proposition 6.3.3. If A ⊆ Nk is representable, then A is Σ-definable.

• Proof. Let A ⊆ N for simplicity and assume that ϕ(v1) represents A.

Let β(x) be the following Σ-formula: β(x) :≡ ∃x∃y

• Num(x, y) ∧ Sub(ϕ, v1, y, z) ∧ ThmN(z)
• .

CLAIM: β(x) defines A. That is, for every n ∈ N, we have n ∈ A ⇐ ⇒ N | = β(n).

Proposition 6.3.3. If A ⊆ Nk is representable, then A is Σ-definable.

• Proof. Let A ⊆ N for simplicity and assume that ϕ(v1) represents A.

Let β(x) be the following Σ-formula: β(x) :≡ ∃x∃y

• Num(x, y) ∧ Sub(ϕ, v1, y, z) ∧ ThmN(z)
• .

CLAIM: β(x) defines A. That is, for every n ∈ N, we have n ∈ A ⇐ ⇒ N | = β(n). First, assume n ∈ A. Since ϕ represents A, we have N ⊢ ϕ(n). Therefore, ϕ(n) ∈ ThmN (by definition of the set ThmN). Since the formula ThmN(z) defines ThmN, it follows that N | = ThmN(ϕ(n)). We have N | = (Num(x, y) ∧ Sub(ϕ, v1, y, z) ∧ ThmN(z))[s] under a vari- able assignment function s : Vars → N with s(x) = n, s(y) = Num(n) = n, s(z) = Sub(ϕ, v1, n) = ϕ(n). Therefore, N | = β(x)[s] ≡ β(n).

Proposition 6.3.3. If A ⊆ Nk is representable, then A is Σ-definable.

• Proof. Let A ⊆ N for simplicity and assume that ϕ(v1) represents A.

Let β(x) be the following Σ-formula: β(x) :≡ ∃x∃y

• Num(x, y) ∧ Sub(ϕ, v1, y, z) ∧ ThmN(z)
• .

CLAIM: β(x) defines A. That is, for every n ∈ N, we have n ∈ A ⇐ ⇒ N | = β(n). For the other direction, assume N | = β(n). Then there exists s : Vars → N with s(x) = n such that N | = (Num(x, y)∧Sub(ϕ, v1, y, z)∧ThmN(z))[s]. Since N | = Num(x, y)[s], it must be that s(y) = Num(s(x)) = Num(n) = n. Since N | = Sub(ϕ, v1, y, z)[s], it must be that s(z) = Sub(ϕ, v1, s(y)) = ϕ(n). Since N | = ThmN(z)[s], we have N | = ThmN(s(z)) ≡ ThmN(ϕ(n)). Since ThmN(z) defines ThmN, it follows that ϕ(n) ∈ ThmN, hence N ⊢ ϕ(n). Finally, since ϕ represents A, we conclude that n ∈ A. Q.E.D.

Proposition 6.3.3. If A ⊆ Nk is representable, then A is Σ-definable.

• Proof. Let A ⊆ N for simplicity and assume that ϕ(v1) represents A.

Let β(x) be the following Σ-formula: β(x) :≡ ∃d “DeductionN(d, Sub(ϕ, v1, Num(x)))”. CLAIM: β(x) defines A. That is, n ∈ A ⇐ ⇒ N | = β(n) for all n ∈ N. First, suppose n ∈ A. Then N ⊢ ϕ(n) (since ϕ represents A). So there exists a deduction (δ1, . . . , δk) of ϕ(n) from N. Let d := δ1, . . . , δk. Then N | = “DeductionN(d, ϕv1

n )”

≡ “DeductionN(d, Sub(ϕ, v1, Num(a))” ≡ ∃y∃z

• Num(n, y) ∧ Sub(ϕ, v1, y, z) ∧ DeductionN(d, z)
• .

Therefore, N | = β(n).

Proposition 6.3.3. If A ⊆ Nk is representable, then A is Σ-definable.

• Proof. Let A ⊆ N for simplicity and assume that ϕ(v1) represents A.

Let β(x) be the following Σ-formula: β(x) :≡ ∃d “DeductionN(d, Sub(ϕ, v1, Num(x)))”. CLAIM: β(x) defines A. That is, n ∈ A ⇐ ⇒ N | = β(n) for all n ∈ N. Conversely, suppose N | = β(n), that is, N | = “DeductionN(d, ϕ(v1))”. Then there exists a deduction of ϕ(v1) from N, hence N ⊢ ϕ(n). Since ϕ represents A, it follows that n ∈ A. Q.E.D.

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• del’s Self-Reference Lemma

Lemma 6.2.2. Let β(x) be an LNT-formula with only x free. Then there is a sentence θ such that N ⊢ θ ↔ β(θ).

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• del’s Self-Reference Lemma

Lemma 6.2.2. Let β(x) be an LNT-formula with only x free. Then there is a sentence θ such that N ⊢ θ ↔ β(θ). Proof Idea. Let’s first define a formula θ with the weaker property that N | = θ ↔ β(θ). We obtain θ by first defining a formula γ(v1) such that, for every formula ϕ(v1), N | = γ(ϕ) ↔ β(ϕ(ϕ)). Once we have such a formula γ(v1), we simply let θ :≡ γ(γ).

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• del’s Self-Reference Lemma

Lemma 6.2.2. Let β(x) be an LNT-formula with only x free. Then there is a sentence θ such that N ⊢ θ ↔ β(θ). Proof Idea. Let’s first define a formula θ with the weaker property that N | = θ ↔ β(θ). We obtain θ by first defining a formula γ(v1) such that, for every formula ϕ(v1), N | = γ(ϕ) ↔ β(ϕ(ϕ)). Definition: γ(v1) :≡ (∃y)(∃z)

• Num(v1, y) ∧ Sub(v1, 8, y, z) ∧ β(z)
• (Note that v1 = 2 = 23 = 8, so 8 ≡ v1 ≡ SSSSSSSS0.)

G¨

• del’s Self-Reference Lemma

Lemma 6.2.2. Let β(x) be an LNT-formula with only x free. Then there is a sentence θ such that N ⊢ θ ↔ β(θ). Proof Idea. Let γ(v1) :≡ (∃y)(∃z)

• Num(v1, y) ∧ Sub(v1, v1, y, z) ∧ β(z)
• θ :≡ γ(γ) :≡ (∃y)(∃z)
• Num(γ, y) ∧ Sub(γ, v1, y, z) ∧ β(z)
• .

G¨

• del’s Self-Reference Lemma

Lemma 6.2.2. Let β(x) be an LNT-formula with only x free. Then there is a sentence θ such that N ⊢ θ ↔ β(θ). Proof Idea. So, we have γ(v1) :≡ (∃y)(∃z)

• Num(v1, y) ∧ Sub(v1, v1, y, z) ∧ β(z)
• θ :≡ γ(γ) :≡ (∃y)(∃z)
• Num(γ, y)
• forces variable assignment y → γ

∧ Sub(γ, v1, y, z) ∧ β(z)

• .

We see that N | = θ ⇐ ⇒ N | = (∃z)

• forces variable assignment z → γ(γ) (= θ)
• Sub(γ, v1, γ, z) ∧ β(z)
• ⇐

⇒ N | = β(θ).

G¨

• del’s Self-Reference Lemma

Lemma 6.2.2. Let β(x) be an LNT-formula with only x free. Then there is a sentence θ such that N ⊢ θ ↔ β(θ). Proof Idea, continued. This formula θ has the property that N | = θ ↔ β(θ). In order to prove the stronger assertion N ⊢ θ ↔ β(θ), we need to replace ∆-formulas Num(x, y) and Sub(x1, x2, x3, y) in θ with: Num∗(x, y) :≡ Num(x, y) ∧ (∀z < y)[¬Num(x, z)] Sub∗(x1, x2, x3, y) :≡ Sub(x1, x2, x3, y) ∧ (∀z < y)[¬Sub(x1, x2, x3, z)]. This lets us use Rosser’s Lemma to eliminate bounded quantifiers in N.