1.4 Theories

Before we start to work with one of the most central notions of mathematical logic – that of a theory – it will be useful to make some observations about countable sets.

Lemma 1.3.

1. A set X is countable iﬀ there is a surjection ψ : N X.

2. If X is countable and Y X then Y is countable.

3. If X and Y are countable, then X Y is countable.

4. If X and Y are countable, then X Y is countable.

5. If X is countable and k 1 then X is countable. ä

6. If Xi is countable for each i N, then Xi is countable.

iÈN

7. If X is countable, then ä Xk is countable.

k ÈN

Proof. 1. If X is countably inﬁnite note that every bijection is a surjection. If X x0,...,xn

is ﬁnite, let ψ i xi for 0 i n and ψ j xn for j n. For the other direction,

let ψ be a surjection, then X is ψ 0 , ψ 1 , ψ 2 ,... and removing duplicates from the

sequence ψ i i¥0 yields a bijective ϕ : X.

2. Obvious if Y is ﬁnite. Otherwise, let ϕ : N X be a bijection and let y0 Y . Deﬁne

ϕn if ϕ n Y ψ : N Y,n y0 otherwise

which is a surjection.

3. Let ϕ : N X and ψ : N Y be bijections. We can obtain a bijection χ : N X Y

by setting

ϕk if n 2k

χn :

ψ k if n 2k 1

We can picture χ in a diagram as

X ? ? ?

where the ﬁrst line represents X in the order given by ϕ and the second line Y in the

order given by ψ and the bijection χ is a path through the inﬁnite graph X Y that touches every vertex exactly once.

4. We can write X Y and a bijection χ : N X Y as a diagram

/ ? / ?

? ?

......

k 1 k

5. By induction on k. If k 1 it is obvious. For k 1 observe that X X X, by assumption X is countable and by induction hypothesis Xk is countable, so by 4 also

Xk 1 is countable.

6. By assumption, there are bijections ϕi : N Xi for every i N. Deﬁne ψ : N N

Xi by ψ i, n ϕi n , then ψ is a surjection. By 4 there is a bijection χ : N N N,

iÈN

ä ä

so ψ χ : N Xi is a surjection hence Xi is countable. È

iÈN i N

7. by 5 and 6.

Given a language L we can assign a number to each symbol that might appear in a formula of L and so we can identify a formula (as well as a term) in L containing k symbols with a k-tuple of natural numbers. By the above result the set of terms as well as the set of formulas of a

given language (and any subset of it, in particular the set of sentences) is countable.

Deﬁnition 1.8. Let Γ be a set of sentences. The deductive closure ClΓ of Γ is deﬁned as

A sentence Γ A .

A set of sentences is called theory if it is deductively closed, i.e. ClT T .

A theory T in a language L is called Henkin theory if for each sentence x A in L there is a

constant cA in L s.t. T x A A x cA .

½ ½ ½ ½ A theory T in a language L is an extension of a theory T in a language L if L L and T T .

A theory T ½ is a conservative extension of a theory T in a language L if it is an extension and

½ if for every L-sentence A: T A implies T A.

A theory T ½ is a Henkin extension of a theory T if it is an extension and a Henkin theory.

Our ﬁrst task will be to construct Henkin theories. These will turn out to be very useful for the

proof of the completeness theorem. Sentences of the form x A A x cA are called Henkin axioms, the cA are called Henkin constants. Let T be a theory in a language L. A ﬁrst naive

attempt to obtaining an Henkin extension of T might be to deﬁne

L : L cA x A sentence in L , and

T : Cl T x A A x cA x A sentence in L

Note that L¦ is countable because L as well as the set of L-sentences are.

Lemma 1.4. If π : Γ A, c is a constant and x a variable which does not appear in π, then

π c x :Γ c x A c x .

Proof. left as exercise (cf. Lemma 1.2).

Lemma 1.5. T ¦ is a conservative extension of T . ¦

Proof. It is obviously an extension. For conservativity suppose T A with A in L. Then

there are Γ T , Henkin axioms H1,...,Hn s.t. Γ,H1,...,Hn A. We proceed by induction

on n. For n 0, T A. For n 0, let Hn xB B x cB . Then

Γ,H1,...,Hn¡1 xB B x cB A.

By Lemma 1.4

Γ,H1,...,Hn¡1 xB B x y A

Γ,H1,...,Hn¡1 y xB B x y A

and using quantiﬁer shiftings

Γ,H1,...,Hn¡1 xB yB x y A

hence

Γ,H1,...,Hn¡1 A.

We obtain T A by induction hypothesis.

Lemma 1.6.

Let T be a theory, deﬁne T0 : T , Tn 1 : Tn and Tω : Ti. Then Tω is a

i¥0

conservative Henkin extension of T .

ä ä

Proof. Note that Ti Ti 1 and hence Cl Ti Cl Ti (see exercises). Therefore ¥

i¥0 i 0

¤ ¤ ¤

Tω Ti Cl Ti Cl Ti Cl Tω

¥ ¥

i¥0 i 0 i 0 is a theory. It is conservative because Tω A implies that there is a k s.t. Tk A and Tk

is conservative over T by induction using Lemma 1.5. Tω is a Henkin theory because for x A

ω A A A ω

in L , x A is in some Lk so c is in Lk 1 and Tk 1 x A A x c so c is in L and

Tω x A A x cA (where Li is the language of Ti).

Deﬁnition 1.9. A theory T in a language L is called complete if for every sentence A in L:

T A or T A. A theory T is called consistent if T .

Lemma 1.7. Every consistent theory T has a consistent complete extension T ½ in the same language.

Proof. Let L be the language of T and let Ai be an enumeration of all sentences in L. Deﬁne

T0 : T

ClTn An if Cl Tn An is consistent

Tn 1 :

Tn otherwise ¤

Tω : Ti

i¥0

Then Tω is consistent for suppose Tω , then there are A1,...,An Tω s.t. A1,...,An

and therefore there is a k s.t. Tk but all Tk are consistent. Furthermore Tω is complete: Let

A be in L, then A Ak for some k. If Cl Tk Ak is consistent, then Tk 1 Ak. Otherwise

ClTk Ak is inconsistent, i.e. Tk, Ak , but then Tk Ak hence Tk Ak by

1.5 Completeness

A central technical tool for the proof of the completeness theorem is the canonical structure of

a theory, sometimes also called term model. Let T be a theory in a language L. We construct

the canonical structure M M, Φ of T as follows. For variable-free terms t,s of T write

t s for T t s. Note that is an equivalence relation, for a term t we write t for its

-equivalence class. The elements of M are the equivalence classes of . We now deﬁne Φ by

1. Φc c for a constant symbol c,

2. Φf t1 ,..., tn f t1,...,tn for a function symbol f, and

3. t1 ,..., tn Φ P iﬀ T P t1,...,tn for a predicate symbol P .

For the above items 2 and 3 to be well-deﬁned we have to verify that the deﬁnition does not

depend on the choice of the representative ti of the equivalence class ti . Suppose we had

chosen diﬀerent representatives s1,...,sn, then, as si ti also T si ti and by the equality

rules of NK we have T f s1,...,sn f t1,...,tn hence f s1,...,sn f t1,...,tn and

T P t1,...,tn iﬀ T P s1,...,sn so Φ is well-deﬁned.

Lemma 1.8. Let T be a consistent complete Henkin theory, M be its canonical structure, A be

a sentence. Then M A iﬀ T A.

Proof. Let M M, Φ . We ﬁrst show Φ t t by induction

1. Φc c by deﬁnition of Φ.

2. Φf t1,...,tn Φ f Φ t1 ,..., Φ tn Φ f t1 ,..., tn f t1,...,tn .

Now we proceed to show M A iﬀ T A by induction on A

1. M P t1,...,tn iff Φ t1 ,..., Φ tn Φ P iff t1 ,..., tn Φ P iff T P t1,...,tn .

2. M s t iff Φ s Φ t iff s t iff s t iff T s t.

3. M A iﬀ M A iﬀ (by IH) T A.

T A implies T A by completeness and T A implies T A by consistency.

4. M A B iﬀ M A or M B iﬀ, by induction hypothesis, T A or T B.

Then T A B because T is deductively closed. If, on the other hand, T A B,

then T A or T B for suppose T A and T B then by completeness T A and T B which contradicts consistency of T .

5. M x A iﬀ there is a t M s.t., for Φ Φ c t for a fresh constant symbol

½ ½ ½ ½

c, M, Φ A x c . Now Φ c Φ t so by applying Lemma 1.1, M, Φ A x c iﬀ

½ ½

M, Φ A x t . But c does not appear in A x t , so M, Φ A x t iﬀ M, Φ A x t

which, by induction hypothesis, is equivalent to T A x t . To sum up, we have proved

M x A iﬀ there is a t s.t. T A x t .

If T A x t then, by T being deductively closed, also T x A. For the other direction,

let T x A, then as T is a Henkin theory, there is a constant cA s.t. T A x cA and

letting t cA we obtain M x A.

The other cases are left as exercises or alternatively follow from normal form theorems.

We have already seen that the set of terms in a countable language is itself countable. The elements of the canonical model of a theory are equivalence classes of terms and therefore a canonical model is countable, i.e. has a countable domain.

Lemma 1.9 (Main Lemma). Every consistent theory T has a countable model.

Proof. By Lemma 1.6 there exists a conservative, and thus consistent, Henkin extension T ½ of

½ ¦

T . By Lemma 1.7 there is a consistent and complete extension T ¦ of T . T is a Henkin theory

½ ¦

too because T ¦ and T are in the same language. Let M be the canonical structure of T , then

¦ ¦ M T by Lemma 1.8 and M T because T is an extension of T .

Theorem 1.2 (Completeness). If A is a valid sentence then A is provable.

Proof. If A is valid then A is unsatisﬁable. So, by Lemma 1.9, Cl A is inconsistent, i.e. there is a proof

i A . .

i A raa