Final Exam Tuesday December 10, 9-12am Closed book exam Cover - - PowerPoint PPT Presentation

Final Exam

• Tuesday December 10, 9-12am
• Closed book exam
• Cover Chapters 1-6 of the textbook.

Might also include a question on quantifier rank or the Ehrenfeucht-Fra¨ ıss´ e game (see Oct 29 slides).

• Greater focus on Chapter 3.3 through 6.
• Won’t ask about material that we didn’t cover (or only barely) cover in
• lecture. For example, no question on the Lowenheim-Skolem Theorems.

Final Exam

• Expect at least one question involving the Compactness Theorem.
• Expect a question on writing down (Σ/Π/∆-)formulas to express specific

properties.

• You don’t need to memorize the axioms of Robinson Arithmetic.
• You won’t be asked to write a deduction. However, by now you should know

the logical axioms and rules of inference.

• Don’t need to memorize the proof of completeness theorem, but good to

understand the principle behind Henkin axioms, etc.

• You should know the sequence-coding function ·. However, you don’t

need to memorize the specific definition of G¨

• del numbers (e.g., t1 + t2 =

13, t1, t2).

Final Exam

• Know the definitions of isomorphism (page 27) and elementary equivalence.
• Know what is a term vs. formula vs. number: t, ϕ, a1, . . . , ak are

numbers: a and ϕ are terms, DeductionN(a, b) is a ∆-sentence, etc.

• Know the definition of Σ/Π/∆-definable sets, (weakly) representable sets,

and recursive/complete/consistent sets of LNT-formulas (today’s lecture).

• Understand the concept of “construction sequences” as a means of building

∆-formulas.

• Know that DeductionN is ∆-definable, ThmN is Σ-definable, etc.
• Know the key definitions from Chapter 1-3: free variables, substitution, |

= and ⊢, statements of Soundness/Deduction/Model Existence/Completeness/ Compactness Theorems.

Key Results

• 1. Rosser’s Lemma: N ⊢ (∀x < a)(x = 0 ∨ x = 1 ∨ · · · ∨ x = a − 1),

N ⊢ [(∀x < a)ϕ(x)] ↔ [ϕ(0) ∧ · · · ∧ ϕ(a − 1)]

• 2. Every Σ-sentence which is true in N is provable from N.
• 3. Every ∆-definable set is representable.
• 4. Church’s Thesis: A set A ⊆ N is (weakly) representable if, and only if, the

question n ∈? A is (semi-)decidable.

• 5. Num, Sub, DeductionN are ∆-definable, via the notion of construction

sequences.

• 6. ThmN is Σ-definable.
• 7. Every representable set is Σ-definable.

Self-Reference Lemma: For every formula β(x), there exists a sentence θ such that N ⊢ θ ↔ β(θ).

Self-Reference Lemma: For every formula β(x), there exists a sentence θ such that N ⊢ θ ↔ β(θ). Summary of Proof: (1) Let γ(v1) :≡ (∃y)(∃z)[Num(v1, y) ∧ Sub(v1, v1, y, z) ∧ β(z)]. This has the property that, for every formula ϕ(v1), N | = γ(ϕ) ↔ β(ϕ(ϕ)).

Self-Reference Lemma: For every formula β(x), there exists a sentence θ such that N ⊢ θ ↔ β(θ). Summary of Proof: (1) Let γ(v1) :≡ (∃y)(∃z)[Num(v1, y) ∧ Sub(v1, v1, y, z) ∧ β(z)]. This has the property that, for every formula ϕ(v1), N | = γ(ϕ) ↔ β(ϕ(ϕ)). (2) Letting θ :≡ γ(γ), we have N | = θ ↔ β(θ).

Self-Reference Lemma: For every formula β(x), there exists a sentence θ such that N ⊢ θ ↔ β(θ). Summary of Proof: (1) Let γ(v1) :≡ (∃y)(∃z)[Num(v1, y) ∧ Sub(v1, v1, y, z) ∧ β(z)]. This has the property that, for every formula ϕ(v1), N | = γ(ϕ) ↔ β(ϕ(ϕ)). (2) Letting θ :≡ γ(γ), we have N | = θ ↔ β(θ). (3) To get the stronger property N ⊢ θ ↔ β(θ), we replace Num(x, y) and Sub(x1, x2, x3, y) in θ with formulas 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)] which strongly represent functions Num and Sub (see book).

Theorem 6.3.10 (Tarski’s Undefinability Theorem, 1936). The set {ϕ : N | = ϕ} of G¨

• del numbers of formulas true in N is not definable.

In other words, truth is undefinable!

• Proof. Toward a contradiction, assume there is a formula α(x) which defines

{ϕ : N | = ϕ}. This mean that, for every formula ϕ, N | = α(ϕ) ⇐ ⇒ N | = ϕ. By the Self-Reference Lemma applied to the formula β(x) :≡ ¬α(x), there is a sentence θ such that N ⊢ θ ↔ β(θ) (and therefore N | = θ ↔ β(θ)). (Think of θ as saying “I am true in N if, and only if, I am false in N.”) This immediately yields a contradiction, as we have N | = θ ⇐ ⇒ N | = β(θ) ⇐ ⇒ N | = α(θ) ⇐ ⇒ N | = θ. Q.E.D.

Theorem 6.3.10 (Tarski’s Undefinability Theorem, 1936). The set {ϕ : N | = ϕ} of G¨

• del numbers of formulas true in N is not definable.

In other words, truth is undefinable!

• Corollary. {ϕ : N |

= ϕ} is not representable. Therefore (assuming Church’s Thesis), there is no computer program which, given an LNT- formula ϕ as input, outputs “true” if N | = ϕ and outputs “false” if N | = ϕ.

Theories (in an arbitrary language L) A theory is a collection of formulas T that is closed under deduction: for every formula ϕ, if T ⊢ ϕ, then ϕ ∈ T. A theory T is consistent if T ⊢ ⊥ (equivalently: if T has a model). A theory T is complete if, for every sentence σ, either σ ∈ T or ¬σ ∈ T.

Theories (in an arbitrary language L) A theory is a collection of formulas T that is closed under deduction: for every formula ϕ, if T ⊢ ϕ, then ϕ ∈ T. A theory T is consistent if T ⊢ ⊥ (equivalently: if T has a model). A theory T is complete if, for every sentence σ, either σ ∈ T or ¬σ ∈ T. Recall: If A is a structure, then the theory of A, written Th(A), is the set

• f formulas that are true in A:

Th(A) = {formulas ϕ : A | = ϕ}. Exercise: (Try on your own!)

• 1. For every structure A, the theory Th(A) is complete and consistent.
• 2. Every complete and consistent theory equals Th(A) for some structure A.

Some Interesting Complete Theories

• Th(N) = Th(N, 0, 1, S, +, ·, E, <) (True Arithmetic, a.k.a., the Theory of

Arithmetic)

• Th(N, 0, 1, +) (Presburger Arithmetic)
• Th(R, 0, 1, +, ·, <) (the Theory of Real Closed Fields)
• Euclidean Geometry: Th(R2, Between, Congruent) where

Between := {(a, b, c) ∈ (R2)3 : b ∈ ac}, Congruent := {(a, b, c, d) ∈ (R2)4 : |ab| = |cd|}. Here ab denotes the line segment between points a, b ∈ R2, and |ab| is the length of ab.

Axioms Notational switch: A will be represent a set of formulas (rather than the universe of a structure, or a subset of Nk) We will regard formulas of A as “axioms” describing a particular theory of interest, such as Th(N) or the theory of linear orders.

Axioms If A is a set of formulas, then the theory of A, written Th(A), is defined by Th(A) = {formulas ϕ : A ⊢ ϕ}. Note that Th(A) is the smallest theory that includes A. A theory T is axiomatized by a set of formulas A if T = Th(A). Example: Let A := { x = y ∨ x < y ∨ y < x, (x < y ∧ y < z) → x < z, ¬(x < x) }. Then A axiomatizes the theory T = {L{<}-formulas that are true in all linear orders}. Non-example: Robinson Arithmetic N does not axiomatize Th(N)

Robinson Arithmetic The eleven axioms of N are: N 1 (∀x)¬[Sx = 0] N 2 (∀x)(∀y)[Sx = Sy → x = y] N 3 (∀x)[x + 0 = x] N 4 (∀x)(∀y)[x + Sy = S(x + y)] N 5 (∀x)[x · 0 = 0] N 6 (∀x)(∀y)[(x · Sy) = (x · y) + x] N 7 (∀x)[xE0 = S0] N 8 (∀x)(∀y)[xE(Sy) = (xEy) · x] N 9 (∀x)¬[x < 0] N 10 (∀x)(∀y)[x < Sy ↔ (x < y ∨ x = y)] N 11 (∀x)(∀y)[x < y ∨ x = y ∨ y < x].

Robinson Arithmetic N is a very nice set of axioms. It is finite! And axioms N1, . . . , N11 are evidently true in N. N is reasonably powerful: It proves every Σ-sentence in Th(N). However, N does not axiomatize Th(N), since (for example) ∀x∀y(x + y = y + x) / ∈ Th(N) \ Th(N).

“Useful” Axiomatization of Th(N) We would like to strengthen N by including more axioms, which allow us prove more theorems in Th(N). Ideally, we would like a set of axioms A which axiomatizes Th(N) (that is, for any formula ϕ, we want N | = ϕ ⇐ ⇒ A ⊢ ϕ). We could take A = Th(N), but this is not a useful axiomatization: How do you recognize if a formula ϕ belongs to A? (It is unclear how to check whether N | = ϕ in finite time...) Question: Is there a “useful” axiomatization of Th(N)? “Useful” could mean finite. But this seems too restrictive; after all, there are infinitely many logical axioms (e.g., x1 = x1, x2 = x2, . . . ), which admits a nice, uniform description. A more reasonable interpretation of “useful”: A should be decidable by an algorithm.

Recursive sets of axioms

• Definition. Let A be a set of axioms of LNT. We say that A is recursive

if the set {α : α ∈ A} is representable.

Recursive sets of axioms

• Definition. Let A be a set of axioms of LNT. We say that A is recursive

if the set {α : α ∈ A} is representable. Recall Church’s Thesis: A set B ⊆ N is representable if, and only if, there exists a computer program which, given n ∈ N as input, eventually outputs “yes” if n ∈ B and eventually outputs “no” if n / ∈ B. Therefore, a set of axioms A is recursive if, and only if, membership in A can be decided algorithmically in finite time (by a computer or, in principle, a human mathematician).

Recursive sets of axioms Lemma 6.3.5. If A is a recursive set of axioms, then the set ThmA := {ϕ : A ⊢ ϕ} is definable by a Σ-formula ThmA(x).

Recursive sets of axioms Lemma 6.3.5. If A is a recursive set of axioms, then the set ThmA := {ϕ : A ⊢ ϕ} is definable by a Σ-formula ThmA(x).

• Proof. Since A is recursive, the set

AxiomA := {α : α ∈ A} is representable and therefore definable by a Σ-formula AxiomA(x). Recall that DeductionN(y, z) is a ∆-formula, which has AxiomN(x) as a ∆- subformula. By replacing AxiomN(x) with AxiomA(x), the result is a Σ- formula DeductionA(y, z) which defines the set DeductionA :=

• (c, a) : c = δ1, . . . , δn and a = ϕ

where (δ1, . . . , δn) is a deduction from A of ϕ

• .

It follows that the set ThmA is defined by Σ-formula ThmA(x) :≡ (∃y)DeductionA(y, x).

Theorem 6.3.6 (G¨

• del’s First Incompleteness Theorem, 1931).

Suppose that A is a consistent and recursive set of axioms in the language

• LNT. Then there is a sentence θ such that N |

= θ but A ⊢ θ.

Theorem 6.3.6 (G¨

• del’s First Incompleteness Theorem, 1931).

Suppose that A is a consistent and recursive set of axioms in the language

• LNT. Then there is a sentence θ such that N |

= θ but A ⊢ θ.

• Proof. If A ⊢ Ni for any of the axioms N1, . . . , N11 of Robinson arithmetic,

then we are done (simply let θ :≡ Ni). So we will assume that A ⊢ N.

Theorem 6.3.6 (G¨

• del’s First Incompleteness Theorem, 1931).

Suppose that A is a consistent and recursive set of axioms in the language

• LNT. Then there is a sentence θ such that N |

= θ but A ⊢ θ.

• Proof. If A ⊢ Ni for any of the axioms N1, . . . , N11 of Robinson arithmetic,

then we are done (simply let θ :≡ Ni). So we will assume that A ⊢ N. Applying the Self-Reference Lemma to the formula β(x) :≡ ¬ThmA(x), we get a sentence θ such that (∗) N ⊢ θ ↔ ¬ThmA(θ). Think of θ as saying “I am true if, and only if, I am not provable from A.”

Theorem 6.3.6 (G¨

• del’s First Incompleteness Theorem, 1931).

Suppose that A is a consistent and recursive set of axioms in the language

• LNT. Then there is a sentence θ such that N |

= θ but A ⊢ θ.

• Proof. If A ⊢ Ni for any of the axioms N1, . . . , N11 of Robinson arithmetic,

then we are done (simply let θ :≡ Ni). So we will assume that A ⊢ N. Applying the Self-Reference Lemma to the formula β(x) :≡ ¬ThmA(x), we get a sentence θ such that (∗) N ⊢ θ ↔ ¬ThmA(θ). Think of θ as saying “I am true if, and only if, I am not provable from A.” Since N | = N, it follows from (∗) that N | = θ ↔ ¬ThmA(θ). Therefore, N | = θ ⇐ ⇒ N | = ¬ThmA(θ) (∗∗) ⇐ ⇒ θ / ∈ ThmA ⇐ ⇒ A ⊢ θ.

Theorem 6.3.6 (G¨

• del’s First Incompleteness Theorem, 1931).

Suppose that A is a consistent and recursive set of axioms in the language

• LNT. Then there is a sentence θ such that N |

= θ but A ⊢ θ. Proof, cont’d. We have: A ⊢ N, N ⊢ θ ↔ ¬ThmA(θ), (∗) N | = θ ⇐ ⇒ N | = ¬ThmA(θ) ⇐ ⇒ A ⊢ θ. (∗∗)

• Claim. It must be the case N |

= θ (hence A ⊢ θ by (∗∗), finishing the proof).

Theorem 6.3.6 (G¨

• del’s First Incompleteness Theorem, 1931).

Suppose that A is a consistent and recursive set of axioms in the language

• LNT. Then there is a sentence θ such that N |

= θ but A ⊢ θ. Proof, cont’d. We have: A ⊢ N, N ⊢ θ ↔ ¬ThmA(θ), (∗) N | = θ ⇐ ⇒ N | = ¬ThmA(θ) ⇐ ⇒ A ⊢ θ. (∗∗)

• Claim. It must be the case N |

= θ (hence A ⊢ θ by (∗∗), finishing the proof). Proof of Claim. Toward a contradiction, assume that N | = θ. Then (∗∗) implies A ⊢ θ.

Theorem 6.3.6 (G¨

• del’s First Incompleteness Theorem, 1931).

Suppose that A is a consistent and recursive set of axioms in the language

• LNT. Then there is a sentence θ such that N |

= θ but A ⊢ θ. Proof, cont’d. We have: A ⊢ N, N ⊢ θ ↔ ¬ThmA(θ), (∗) N | = θ ⇐ ⇒ N | = ¬ThmA(θ) ⇐ ⇒ A ⊢ θ. (∗∗)

• Claim. It must be the case N |

= θ (hence A ⊢ θ by (∗∗), finishing the proof). Proof of Claim. Toward a contradiction, assume that N | = θ. Then (∗∗) implies A ⊢ θ. (∗∗) also implies N | = ThmA(θ). Since ThmA(θ) is Σ-sentence which is true in N, we have N ⊢ ThmA(θ) (Proposition 5.3.13). By (∗) and the (PC) rule, it follows that N ⊢ ¬θ. Since A ⊢ N, we have A ⊢ ¬θ.

Theorem 6.3.6 (G¨

• del’s First Incompleteness Theorem, 1931).

Suppose that A is a consistent and recursive set of axioms in the language

• LNT. Then there is a sentence θ such that N |

= θ but A ⊢ θ. Proof, cont’d. We have: A ⊢ N, N ⊢ θ ↔ ¬ThmA(θ), (∗) N | = θ ⇐ ⇒ N | = ¬ThmA(θ) ⇐ ⇒ A ⊢ θ. (∗∗)

• Claim. It must be the case N |

= θ (hence A ⊢ θ by (∗∗), finishing the proof). Proof of Claim. Toward a contradiction, assume that N | = θ. Then (∗∗) implies A ⊢ θ. (∗∗) also implies N | = ThmA(θ). Since ThmA(θ) is Σ-sentence which is true in N, we have N ⊢ ThmA(θ) (Proposition 5.3.13). By (∗) and the (PC) rule, it follows that N ⊢ ¬θ. Since A ⊢ N, we have A ⊢ ¬θ. We have now shown that A ⊢ θ and A ⊢ ¬θ. But this means that A ⊢ ⊥, contradicting our initial assumption that A is consistent. Q.E.D.

Self-Reference Lemma. If β(x) is an LNT-formula with only x free, then there is a sentence θ such that N ⊢ θ ↔ β(θ). Tarski’s Undefinability Theorem. The set {ϕ : N | = ϕ} of G¨

• del

numbers of formulas true in N is not definable.

• Idea. Toward a contradiction, assume α(x) defines {ϕ : N |

= ϕ}. By Self- Reference Lemma applied to β(x) :≡ ¬α(x), there is a sentence θ such that N | = θ ↔ ¬α(θ). This yields a contradiction: N | = θ ⇐ ⇒ N | = ¬β(θ) ⇐ ⇒ N | = θ. G¨

• del’s 1st Incompleteness Theorem. Suppose that A is a consistent

and recursive set of axioms in the language LNT. Then there is a sentence θ such that N | = θ but A ⊢ θ.

• Idea. If A ⊢ N, we’re done. If A ⊢ N, let β(x) :≡ ¬ThmA(x). By Self-

Reference Lemma, there is a sentence θ such that N ⊢ θ ↔ ¬ThmA(θ). This forces N | = θ and A ⊢ θ (otherwise we get a contradiction).

Theorem 6.4.5 (Rosser’s Theorem). If A is a set of LNT-axioms that is recursive, consistent, and extends N, then A is incomplete. In other words, Robinson arithmetic N is not “recursively completable”.

Theorem 6.4.5 (Rosser’s Theorem). If A is a set of LNT-axioms that is recursive, consistent, and extends N, then A is incomplete. Proof Idea (Rosser’s Trick). Use the Self-Reference Lemma to construc- tion a sentence θ which expresses: “I am true if, and only if, for every deduction

• f A ⊢ θ, there is a shorter deduction of A ⊢ ¬θ.”

Obs: If a := θ, then ¬θ = 1, θ = 223θ+1 = 4 · 3a+1. Let β(x) :≡ (∀y)

• DeductionA(y, x) → (∃z < y)DeductionA(z, 4 · 3

Sx)

• .

By the Self-Reference Lemma, we get a sentence θ such that N ⊢ θ ↔ β(θ) ≡ θ ↔ (∀y)

• DeductionA(y, θ) → (∃z < y)DeductionA(z, ¬θ)
• .

Theorem 6.4.5 (Rosser’s Theorem). If A is a set of LNT-axioms that is recursive, consistent, and extends N, then A is incomplete. Proof Idea (Rosser’s Trick). Use the Self-Reference Lemma to construc- tion a sentence θ which expresses: “I am true if, and only if, for every deduction

• f A ⊢ θ, there is a shorter deduction of A ⊢ ¬θ.”

Obs: If a := θ, then ¬θ = 1, θ = 223θ+1 = 4 · 3a+1. Let β(x) :≡ (∀y)

• DeductionA(y, x) → (∃z < y)DeductionA(z, 4 · 3

Sx)

• .

By the Self-Reference Lemma, we get a sentence θ such that N ⊢ θ ↔ β(θ) ≡ θ ↔ (∀y)

• DeductionA(y, θ) → (∃z < y)DeductionA(z, ¬θ)
• .

We get a contradiction if we assume A ⊢ θ, and we get a contradiction if we assume A ⊢ ¬θ. Therefore, A ⊢ θ and A ⊢ ¬θ; hence A is incomplete. Q.E.D.

Theorem 6.4.5 (Rosser’s Theorem). If A is a set of LNT-axioms that is recursive, consistent, and extends N, then A is incomplete. Proof Idea (Rosser’s Trick). Use the Self-Reference Lemma to construc- tion a sentence θ which expresses: “I am true if, and only if, for every deduction

• f A ⊢ θ, there is a shorter deduction of A ⊢ ¬θ.”

Obs: If a := θ, then ¬θ = 1, θ = 223θ+1 = 4 · 3a+1. Let β(x) :≡ (∀y)

• DeductionA(y, x) → (∃z < y)DeductionA(z, 4 · 3

Sx)

• .

By the Self-Reference Lemma, we get a sentence θ such that N ⊢ θ ↔ β(θ) ≡ θ ↔ (∀y)

• DeductionA(y, θ) → (∃z < y)DeductionA(z, ¬θ)
• .

We get a contradiction if we assume A ⊢ θ, and we get a contradiction if we assume A ⊢ ¬θ. Therefore, A ⊢ θ and A ⊢ ¬θ; hence A is incomplete. Q.E.D.

• Remark. Either N |

= θ or N | = ¬θ. Either way, we get a sentence true in N but not provable from A. Thus, Rosser’s Theorem ⇒ 1st Incompleteness Theorem.