please ll the course evaluation no tutorial next week o ce ... · final lecture next tuesday { will...

Reminder

Please fill the course evaluation

No tutorial next week

Office hours as usual on Thursday April 5 and 12

Final lecture next Tuesday – will discuss final exam, quick overview of 2ndIncompleteness Theorem, review of Ehrenfeucht-Fraisse games

Midterm

Pick up at end of lecture today

Midterm 1 average (among currently enrolled students): 74.5

Midterm 2 average: 61.5

Distribution of marks

90–98: 6

80–89: 10

70–79: 12

60–69: 16

50–59: 9

40–49: 13

Common Mistakes

• A set of sentences Σ is complete if, for every sentence θ, either Σ ` θ orΣ ` ¬θ.Common incorrect answer: “Σ is complete if Σ |= ϕ ⇒ Σ ` ϕ.” This isthe Completeness Theorem, which says our system of deductions is com-plete (different usage of “complete”, meaning every logical implication hasa deduction).

• The term x+y is substitutable for x in the formula (x < y) ∨ (∃x)[(∀y)(y < Sx)].It is not substitutable for x in the subformula (∀y)(y < Sx).

• The axioms of Robinson arithmetic are not valid LNT -sentences: they arenot true in every LNT -structure.They are of course true in the natural numbers N.

• The set N

∆-definable sets and functions

Using an appropriate notion of “construction sequence”, we get a ∆-formulaNum(x, y) which defines the function

Num(a) := p a q.


Using an appropriate notion of “construction sequence”, we get a ∆-formulaNum(x, y) which defines the function

Num(a) := p a q.

This means:

• N |= Num(a, b) for all (a, b) ∈ N2 such that b = p a q• N |= ¬Num(a, b) for all (a, b) ∈ N2 such that b 6= p a q


Similarly (by a more complicated “construction sequence”), there is a ∆-formulaSub(x1, x2, x3, y) which define the function

Sub(pϕq, pxq, ptq) := pϕxt q.


Similarly (by a more complicated “construction sequence”), there is a ∆-formulaSub(x1, x2, x3, y) which define the function

Sub(pϕq, pxq, ptq) := pϕxt q.

This means: for all (a, b, c, d) ∈ N4,

• N |= Sub(a, b, c, d) if a = pϕq and b = pxq and c = ptq and d = pϕxt q forsome formula ϕ and variable symbol x and term t

• N |= ¬Sub(a, b, c, d) otherwise.


Using ∆-formulas Num(x, y) and Sub(x1, x2, x3, y) (among other useful ∆-formulas such as Free and Substitutable)), we get ∆-formulas defining sets:

LogicalAxiom :={pϕq : ϕ is a logical axiom

}AxiomN :=

{pN1q, . . . , pN11q

},

RuleOfInf :={

(c, a) : c = 〈pγ1q, . . . , pγnq〉 and a = pϕqwhere ({γ1, . . . , γn}, ϕ) is a rule of inference

}.

Finally, we get a ∆-formula DeductionN(y, z) which defines the set

DeductionN :={

(c, a) : c = 〈pδ1q, . . . , pδ1q〉 and a = pϕqwhere (δ1, . . . , δn) is a deduction from N of ϕ

}.

The Σ-formula ThmN(x)

The set

ThmN := {pϕq : N ` ϕ}

is defined by Σ-formula

ThmN(x) :≡ (∃y)DeductionN(y, x).


The set




This means: for every a ∈ N,

• N |= Thm(a) if a = pϕq for some formula ϕ such that N ` ϕ,

(In this case, N ` Thm(a) since N proves every Σ-sentence which is truein N by Proposition 5.3.13.)

• N |= ¬Thm(a) otherwise.

(We cannot conclude that N ` ¬Thm(a) since ¬Thm(a) is (equivalentto) a Π-sentence.)


The set




There is no obvious way to rewrite ThmN(x) as a ∆-sentence (in fact, this isimpossible). For instance, we cannot replace (∃y) with (∃y < xxx), since thiswould imply that every formula of length ` provable by ϕ has a deduction of

length < ```

(which is false).

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. (With-out loss of generality, we take v1 to be the free variable of ϕ.) Let β(x) be thefollowing Σ-formula:

β(x) :≡ ∃y∃z∃w[

Num(x, y) ∧ Sub(pϕq, pv1q, y, z) ∧ DeductionN(w, z)︸︷︷︸call this subformula γ(x, y, z, w)

].

We claim that β(x) defines A.



β(x) :≡ ∃y∃z∃w[


].

Assume a ∈ A. Then N ` ϕ(a) (since N represents A). So there exists adeduction (δ1, . . . , δn) of ϕ(a) from N .

Now consider the following variable assignment:

y 7→ Num(a), z 7→ pϕ(a)q, w 7→ 〈pδ1q, . . . , pδnq〉.

By definition of formulas Num, Sub and DeductionN , we have

N |= γ(a,Num(a), pϕ(a)q, 〈pδ1q, . . . , pδnq〉).

This shows that N |= β(a).



β(x) :≡ ∃y∃z∃w[


].

Conversely, assume a ∈ N such that N |= β(a). Then there exist b, c, d ∈ Nsuch that N |= γ(a, b, c, d). This means that

b = Num(a), c = pϕ(a)q, d = 〈pδ1q, . . . , pδnq〉.

for some formula ϕ and deduction (δ1, . . . , δn) of ϕ(a) from N .

Therefore, N ` ϕ(a). It follows that a ∈ A (again using the fact that ϕrepresents A).

Q.E.D.

Gödel’s Self-Reference Lemma

Lemma 6.2.2. Let β(x) be an LNT -formula with only x free. Then thereis a sentence θ such that

N ` θ ↔ β(pθq).




Proof Idea. Let

γ(v1) :≡ (∃y)(∃z)[Num(v1, y) ∧ Sub(v1, pv1q, y, z) ∧ β(z)

]θ :≡ γ(pγq) :≡ (∃y)(∃z)

[Num(pγq, y) ∧ Sub(pγq, pv1q, y, z) ∧ β(z)

].




Proof Idea. Let

γ(v1) :≡ (∃y)(∃z)[Num(v1, y) ∧ Sub(v1, pv1q, y, z) ∧ β(z)

]θ :≡ γ(pγq) :≡ (∃y)(∃z)

[Num(pγq, y)︸︷︷︸

forces variable assignment y 7→ ppγqq

∧ Sub(pγq, pv1q, y, z) ∧ β(z)].

We can see that

N |= θ ⇐⇒ N |= (∃z)[forces variable assignment z 7→ pγ(pγq)q (= pθq)︷︸︸︷

Sub(pγq, pv1q, ppγqq, z) ∧ β(z)]⇐⇒ N |= β(pθq).




Proof Idea, continued. We just showed that

N |= θ ↔ β(pθq).

In order to prove the stronger assertion

N ` θ ↔ β(pθq),

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 .

Recursive sets of axioms

We have seen that N is an incomplete set of axioms for Th(N): it cannot provebasic theorems like (∀x)(∀y)[x + y = y + x].

We would like to strengthen N by including more axioms, which allow us provemore theorems in Th(N). Ideally, we would like a set of axioms A which is acomplete and consistent axiomatization of Th(N): that is, ϕ ∈ Th(N) ⇐⇒A ` ϕ.


We have seen that N is an incomplete set of axioms for Th(N): it cannot provebasic theorems like (∀x)(∀y)[x + y = y + x].

We would like to strengthen N by including more axioms, which allow us provemore theorems in Th(N). Ideally, we would like a set of axioms A which is acomplete and consistent axiomatization of Th(N): that is, ϕ ∈ Th(N) ⇐⇒A ` ϕ.

Note that Th(N) is a consistent and complete axiomatization of itself. However,Th(N) is useless—to human mathematicians—as a set of axioms, because apriori we have no way of recognizing which LNT -sentences belong to Th(N).

For a set of axioms A to be useful, we require a way of recognizing whichsentences belong to A. (Ideally, our choice of axioms A should be a minimalset of evidently true assertions about N, from which any reasonable questionarising in number theory can be proved or refuted.)


Definition. Let A be a set of axioms of LNT . We say that A is recursiveif the set {pαq : α ∈ A} is representable.


Definition. Let A be a set of axioms of LNT . We say that A is recursiveif the set {pαq : α ∈ A} is representable.

Recall that a subset B ⊆ N is representable if, and only if, there exists analgorithm (say, as formalized by a Turing machine) which, given n ∈ N asinput, determines in finite time whether or not n belongs to B.

Therefore, a set of axioms A is recursive if, and only if, membership in A can bedecided algorithmically (by a computer or in principle a human mathematician).


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


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

Proof. Since A is recursive, the set

AxiomA := {pαq : 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 = 〈pδ1q, . . . , pδ1q〉 and a = pϕqwhere (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 (Gödel’s First Incompleteness Theorem, 1931).Suppose that A is a consistent and recursive set of axioms in the languageLNT . Then there is a sentence θ such that N |= θ but A 6` θ.


Proof. If A 6` 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), weget a sentence θ such that

(∗) N ` θ ↔ ¬ThmA(pθq).

Think of θ as saying “I am true if, and only if, I am not provable from A.”



Applying the Self-Reference Lemma to the formula β(x) :≡ ¬ThmA(x), weget a sentence θ such that

(∗) N ` θ ↔ ¬ThmA(pθq).

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(pθq). Therefore,

N |= θ ⇐⇒ N |= ¬ThmA(pθq)(∗∗)⇐⇒ pθq /∈ ThmA⇐⇒ A 6` θ.


Proof, cont’d. We have: A ` N ,

N ` θ ↔ ¬ThmA(pθq),(∗)N |= θ ⇐⇒ N |= ¬ThmA(pθq) ⇐⇒ A 6` θ.(∗∗)

Claim. It must be the case N |= θ (hence A 6` θ by (∗∗), finishing the proof).





Proof of Claim. Toward a contradiction, assume that N 6|= θ. Then (∗∗)implies A ` θ.






(∗∗) also implies N |= ThmA(pθq). Since ThmA(pθq) is Σ-sentence which istrue in N, we have N ` ThmA(pθq) (Proposition 5.3.13). By (∗) and the (PC)rule, it follows that N ` ¬θ. Since A ` N , we have A ` ¬θ.






(∗∗) also implies N |= ThmA(pθq). Since ThmA(pθq) is Σ-sentence which istrue in N, we have N ` ThmA(pθq) (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.

Theorem 6.3.10 (Tarski’s Undefinability Theorem, 1936). The set{pϕq : N |= ϕ} of Gödel numbers of formulas true in N is not definable.

In other words, truth is undefinable!



Proof. Toward a contradiction, assume there is a formula β(x) which defines{pϕq : N |= ϕ}. This mean that, for every formula α,

N |= β(pαq) ⇐⇒ pαq ∈ {pϕq : N |= ϕ} ⇐⇒ N |= α.





By the Self-Reference Lemma applied to the formula ¬β(x), there is a sentenceθ such that

N ` θ ↔ ¬β(pθq) (and therefore N |= θ ↔ ¬β(pθq)).

Think of θ as saying “I am true if, and only if, I am false.”





By the Self-Reference Lemma applied to the formula ¬β(x), there is a sentenceθ such that

N ` θ ↔ ¬β(pθq) (and therefore N |= θ ↔ ¬β(pθq)).

Think of θ as saying “I am true if, and only if, I am false.”

This immediately yields a contradiction, as we have

N |= θ ⇐⇒ N |= ¬β(pθq) ⇐⇒ N 6|= θ.

Q.E.D.

Theorem 6.4.5 (Rosser’s Theorem). If A is a set of LNT -axioms thatis recursive, consistent, and extends N , then A is incomplete.

In other words, Robinson arithmetic N is not “recursively completable”.



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 deductionof A ` θ, there is a shorter deduction of A ` ¬θ.”

Obs: If a := pθq, then p¬θq = 〈1, pθq〉 = 223pθq+1 = 4 · 3a+1.

Let β(x) be the formula

β(x) :≡ (∀y)[DeductionA(y, x)→ (∃z < y)DeductionA(z, 4 · 3

Sx)].



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 deductionof A ` θ, there is a shorter deduction of A ` ¬θ.”

Obs: If a := pθq, then p¬θq = 〈1, pθq〉 = 223pθq+1 = 4 · 3a+1.

Let β(x) be the formula

β(x) :≡ (∀y)[DeductionA(y, x)→ (∃z < y)DeductionA(z, 4 · 3

Sx)].

By Self-Reference Lemma, there is a sentence θ such that


If we assume A ` θ, we get a contradiction. If we assume A ` ¬θ, we get acontradiction. Therefore, A neither proves nor refutes θ; so A is incomplete.

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