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.
∆-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.
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
∆-definable sets and functions
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.
∆-definable sets and functions
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.
∆-definable sets and functions
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 Σ-formula ThmN(x)
The set
ThmN := {pϕq : N ` ϕ}
is defined by Σ-formula
ThmN(x) :≡ (∃y)DeductionN(y, x).
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 Σ-formula ThmN(x)
The set
ThmN := {pϕq : N ` ϕ}
is defined by Σ-formula
ThmN(x) :≡ (∃y)DeductionN(y, x).
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.
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)
].
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).
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)
].
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).
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)
].
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)︸ ︷︷ ︸
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).
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, 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.
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.
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.)
Recursive sets of axioms
Definition. Let A be a set of axioms of LNT . We say that A is recursiveif the set {pαq : α ∈ A} is representable.
Recursive sets of axioms
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).
Recursive sets of axioms
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).
Recursive sets of axioms
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.
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 .
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.”
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.”
Since N |= N , it follows from (∗) that N |= θ ↔ ¬ThmA(pθq). Therefore,
N |= θ ⇐⇒ N |= ¬ThmA(pθq)(∗∗)⇐⇒ pθq /∈ ThmA⇐⇒ A 6` θ.
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, 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).
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, 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.
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, 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.
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, 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.
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!
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.
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.”
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.”
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”.
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)].
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)].
By Self-Reference Lemma, there is a sentence θ such that
N ` θ ↔ β(pθq).
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.