Hedman. A First Course in Logic, 2004 (Oxford)

We now define the VR-theory TRG. This is the theory of graphs that model ρm for all m N. That is, TRG is the VR-theory axiomatized by:

x¬R(x, x),

x y(R(x, y) ↔ R(y, x)),

x y¬(x = y), and

ρm for each m N.

Let ∆ be a finite subset of this infinite set of sentences. By Lemma 5.35, ∆ is satisfied by a preponderance of the finite graphs of size n for su ciently large n. Since every finite subset is satisfiable, TRG is satisfiable by compactness. So TRG is indeed a theory. We show that it is a complete theory.

Proposition 5.36 TRG is 0-categorical.

Proof Let M and N be two denumerable models of TRG. Let UM and UN denote the sets of vertices of M and N respectively. Enumerate these sets as follows:

UM = {a1, a2, a3, . . .} and UN = {b1, b2, b3, . . .}.

We construct an isomorphism f : M → N using a back-and-forth argument.

Step 1: Let f (a1) = b1.

Step (n + 1): (Part a) Let An be the set of vertices in UM for which f has been defined in some previous step. Let j be least such that aj is not in An. Since N |= ρm for arbitrarily large m, there exists a vertex f (aj ) such that N |= R(f (ai), f (aj )) if and only if M |= R(ai, aj ) for any ai An.

(Part b) Let Bn = {f (ai)|ai An} {f (aj )}. Let j be least such that bj is not in Bn. By the same argument as in part a, we can find f −1(bj ) as desired.

The function f defined in this manner is a one-to-one function from M onto N that preserves the edge relation R. It follows that f is an isomorphism and TRG is 0-categorical.

Definition 5.37 The random graph, denoted by GR, is the unique countable model of TRG.

Proposition 5.38 Any finite graph can be embedded into any model of TRG.

Proof Let G be a finite graph. Let M be an arbitrary model of TRG. We show that G embeds into M by induction on n = |G|. Clearly this is true if n = 1. Suppose that any graph of size m embeds into M for some m N. Let G be a graph having vertices {v1, . . . , vm+1}. By our induction hypothesis, the substructure G of G having vertices {v1, . . . , vm} can be embedded into M . Let

f : G → M denote this embedding. Since M models ρk for arbitrarily large k, there exists a vertex f (an+1) of M such that M |= R(f (ai), f (an+1)) if and only if G |= R(ai, an+1) for i = 1, . . . , n. Thus G is embedded into M . By induction, any finite graph can be embedded into M .

Proposition 5.39 TRG is complete.

Proof It follows from the previous proposition that TRG has only infinite models. By Proposition 5.36, TRG is 0-categorical. By Proposition 5.15, TRG is complete.

Theorem 5.40 (0–1 Law for Graphs) For every VR-sentence θ, either limn→∞ Pn(θ) = 0 or limn→∞ Pn(θ) = 1.

Proof Recall the axiomatization that was given for TRG. By Lemma 5.35, limn→∞ Pn(ϕ) = 1 for each sentence ϕ in this axiomatization. It follows that limn→∞ Pn(ϕ) = 1 for every sentence ϕ in TRG. Since TRG is complete, either TRG θ or TRG ¬θ for every VR-sentence θ. It follows that either limn→∞ Pn(θ) = 1 or limn→∞ Pn(¬θ) = 1.

This result can be generalized. A vocabulary is relational if it contains no functions. So relational vocabularies may contain constants as well as relations. Let Mn be the set of all V-structures having underlying set {1, 2, 3, . . . , n}. Let PVn(θ) be the number of structures in Mn that model θ divided by |Mn|.

Theorem 5.41 (0–1 Law for Relations) Let V be a finite relational vocabulary. For any V-sentence θ, either limn→∞ PVn(θ) = 0 or limn→∞ PVn(θ) = 1.

5.5 Quantifier elimination

Suppose that we want to analyze a given first-order structure M . We could begin by trying to find an axiomatization for T h(M ). Suppose we have accomplished this and T h(M ) is decidable. Then, for any sentence ϕ in the vocabulary of M , we can determine whether M |= ϕ or M |= ¬ϕ. However, understanding the theory T h(M ) is only a first step toward understanding the structure M . To analyze M further, one must be familiar with the definable subsets of the structure.

For example, suppose that we are presented with a rather complicated graph G. We are given the set of vertices {v1, v2, v3, . . .} along with the set of all pairs of vertices that share edges in G. Suppose too that we are given a decidable axiomatization of T h(G). Then for any VR-sentence ϕ, we can determine whether or not ϕ holds in G. In some sense, this data represents all there is to know about the structure G. But suppose we want to determine which pairs of vertices (x1, x2) satisfy the VR-formula ψ(x1, x2) defined by

y z u v(R(x1, y) R(y, z) R(z, u) R(u, v) R(v, x2)).

If neither x1 x2ψ(x1, x2) nor x1 x2¬ψ(x1, x2) hold in G, then it may be a di cult task to determine whether or not a given pair of vertices satisfies this formula. In the terminology of Section 4.5.2, ψ(x1, x2) is a 4 formula. If you find the formula ψ(x1, x2) easy to comprehend, then consider a 23 formula or a 45 formula. In Chapter 9, we shall introduce a technique that helps us get a handle on such complicated formulas (pebble games). In the present section, we study a property that allows us to utterly avoid them.

Definition 5.42 A V-theory T has quantifier elimination if every V-formula ϕ(x1, . . . , xn) (for n N) there exists a quantifier-free V-formula ψ(x1, . . . , xn) such that T ϕ(x1, . . . , xn) ↔ ψ(x1, . . . , xn).

Quantifier elimination is a purely syntactic property that greatly facilitates the study of certain mathematical structures. If a V-theory has this property, then every V-definable subset of every model is defined by a quantifier-free formula.

For example, suppose G is a graph that has quantifier elimination. Since all vertices of a graph satisfy the same quantifier-free formulas (namely x = x and ¬R(x, x)), any VR-formula in one free variable either holds for all vertices or no vertices of G. For a pair of distinct vertices x1 and x2 of G, there are two possibilities: either R(x1, x2) or ¬R(x1, x2) holds in G. In particular, we can determine whether or not the above 4 formula ψ(x1, x2) holds merely by checking whether or not x1 and x2 share an edge. One of the following two sentences must be in the VR-theory of G:

x1 x2(R(x1, x2) → ψ(x1, x2)) or x1 x2(R(x1, x2) → ¬ψ(x1, x2)).

Likewise, any graph having quantifier elimination must model either

x1 x2(¬R(x1, x2) → ψ(x1, x2)) or x1 x2(¬R(x1, x2) → ¬ψ(x1, x2)).

The goal of this section is to formulate methods that determine whether or not a given complete theory has quantifier elimination.

5.5.1 Finite relational vocabularies. Let T be a complete theory, and suppose that we want to determine whether or not T has quantifier elimination. We make two initial observations.

•Theorem 4.49 provides a su cient criterion for a formula to be T -equivalent to a quantifier-free formula.

•To show that T has quantifier elimination, it su ces to check this criterion only for existential formulas having only one occurences of “ .”

We elaborate and verify the latter point.

Let T be a V-theory. To show that T has quantifier elimination, we must show that ϕ(¯x) is T -equivalent to a quantifier-free formula for every V-formula

ϕ(¯x) having at least one free variable. One way to do this is to proceed by induction on the complexity of ϕ(¯x). If ϕ(¯x) is quantifier-free, then there is nothing to show. Suppose that both ψ and θ are T -equivalent to quantifier-free formulas. If ϕ(¯x) is T -equivalent to either of these formulas, their negations, or their conjunction, then ϕ(¯x) is T -equivalent to a quantifier-free formula. Now suppose ϕ(¯x) is equivalent to yψ(¯x, y). To show that T has quantifier elimination, it su ces to show that this formula is T -equivalent to a quantifier-free formula. In this way, the problem of showing that T has quantifier elimination reduces to the problem of showing that formulas of the form yψ(¯x, y) are T -equivalent to quantifier-free formulas.

Proposition 5.43 A V-theory T has quantifier elimination if and only if for every quantifier-free V-formula ϕ(x1, . . . , xn, y) (for n N), there exists a quantifierfree V-formula ψ(x1, . . . , xn) such that T yϕ(x1, . . . , xn, y) ↔ ψ(x1, . . . , xn).

Proof Suppose that yϕ(¯x, y) is T -equivalent to a quantifier-free formula for every quantifier-free V-formula ϕ having at least two free variables. Then we can show that every V-formula θ is T -equivalent to a quantifier-free formula by induction on the complexity of θ as in the preceeding paragraph. Conversely, if T has quantifier elimination, then yϕ(¯x, y), like every V-formula, is T equivalent to a quantifier-free formula.

So to eliminate all of the quantifiers from a formula like

y z u v(R(x1, y) R(y, z) R(z, u) R(u, v) R(v, x2)),

we need only be able to eliminate one occurrence of the quantifier at a time. For complete T , Theorem 4.49 gives us a criterion for determining whether a given formula is T -equivalent to a quantifier free formula. This yields a method for showing quantifier elimination. We first consider theories that have finite relational vocabularies. These vocabularies are particularly simple because of the following fact.

Proposition 5.44 The vocabulary V is finite and relational if and only if there are only finitely many atomic V-formulas.

Proof If V contains a function f , then we have the atomic formulas f (¯x) = y, f (f (¯x)) = y, f (f (f (¯x)) = y, and so forth.

Proposition 5.45 Let T be a complete theory in a finite relational vocabulary. The following are equivalent.

(i)T has quantifier-elimination.

(ii)For any model M of T and any n N, if (a1, . . . , an) and (b1, . . . , bn) are n-tuples of UM that satisfy the same atomic formulas in M ,

then for any an+1 UM there exists bn+1 UM such that (a1, . . . , an, an+1), and (b1, . . . , bn, bn+1) satisfy the same atomic formulas in M (where UM denotes the underlying set of M ).

Proof Suppose first that T has quantifier elimination.

If (a1, . . . , an) and (b1, . . . , bn) satisfy the same atomic formulas, then they satisfy the same quantifier-free formulas. This can be shown by induction on the complexity of a given quantifier-free formula. Given an+1 UM , let Φ(x1, . . . , xn+1) be the conjunction of all of the atomic and negated atomic formulas that hold of (a1, . . . , an, an+1) in M . Such a formula Φ exists since there are only finitely many atomic formulas. By quantifier elimination,yΦ(x1, . . . , xn, y) is T -equivalent to a quantifier-free formula θ(x1, . . . , xn). Since M models θ(a1, . . . , an), M also models θ(b1, . . . , bn). It follows that M |= yΦ(b1, . . . , bn, y). Let bn+1 UM be such that M |= Φ(b1, . . . , bn, bn+1).

Conversely, suppose that (ii) holds. Let ϕ(x1, . . . , xn, xn+1) be a quantifier-

equivalent to a quantifier-free formula. By Proposition 5.43, T has quantifier elimination.

Example 5.46 Recall the VS -structure ZS = (Z|S) from Example 5.51. This structure interprets the binary relation S as the successor relation on the integers. As was pointed out in Example 4.48, the ordered pairs (0, 2) and (4, 7) satisfy the same atomic VS -formulas. Let TS = T h(ZS ). If TS had quantifier elimination, then, by condition (ii) of Proposition 5.45, for every integer x there would exist an integer y such that (0, 2, x) and (4, 7, y) satisfy the same atomic formulas. To show that this is not the case, let x = 1. Then both S(0, x) and S(x, 2) hold. Clearly, there is no y that bears these relations to 4 and 7. We conclude that TS does not have quantifier elimination. In particular, the formula z(S(x, z) S(z, y)) is not TS -equivalent to a quantifier-free formula (this was also shown in Example 4.48).

Example 5.47 We show that TDLOE does not have quantifier elimination. The only two atomic V<-formulas are x < y and x = y. Any two n-tuples of elements listed in ascending order will satisfy the same atomic formulas. In particular, this is true if n = 1. Consider the two elements 0 and 0.01 from the underlying set of Q[0,1]. These two elements satisfy the same atomic V<-formulas in Q[0,1]. However, since 0 is the smallest element, there is no y so that (0, y) satisfies the same atomic formulas as (0.01, 0.001). By Proposition 5.45, TDLOE does not have quantifier elimination. In particular, the formula y(y < x) is not TDLOE -equivalent to a quantifier-free formula.

Using condition (ii) of Proposition 5.45, we can quickly show that some theories do not have quantifier elimination as in the previous examples. To show that a theory T does have quantifier elimination, (ii) requires us to consider all pairs of tuples from all models of T . If T is complete and has a finite relational vocabulary, then this condition can be simplified as the following corollary states.

Corollary 5.48 Let T be a complete theory. If T has a finite relational vocabulary, then T has quantifier elimination if and only if condition (ii) from Proposition 5.45 holds in some model M of T .

Proof This follows from the assumption that T is complete. Let Φ and θ be as in the proof of Proposition 5.45. Let M be some model of T . If M modelsyΦ(x1, . . . , xn, y) ↔ θ(x1, . . . , xn), then, since T is complete, so does every model of T .

Proposition 5.49 The theory TDLO of dense linear orders without endpoints has quantifier elimination.

Proof By the previous corollary, it su ces to verify condition (ii) of Proposition 5.45 for only one model. Let Q< be the V<-structure that interprets < as the usual order on the rational numbers.

¯

Let a¯ = (a1, . . . , an) and b = (b1, .., bn) be two n-tuples of rational numbers. With no loss of generality, we may assume that a1 < a2 < · · · < an. Suppose that

¯ V · · ·

a¯ and b satisfy the same atomic <-formulas. Then we have b1 < b2 < < bn. Let an+1 be any rational number. We must show that there exists a rational number bn+1 so that (b1, . . . , bn, bn+1) satisfies the same atomic V<-formulas as (a1, . . . , an, an+1).

There are four cases:

If an+1 = ai for some i = 1, . . . , n, then we can just let bn+1 = bi.

If an+1 < ai for each i, then, since Q has no smallest element, we can find bn+1 Q that is smaller than each bn+1.

Likewise, if an+1 is greater than each ai, then we can find bn+1 Q that is greater than each bi.

Otherwise, ai < an+1 < ai+1 for some i. Since Q is dense, we can find bn+1 Q between bi and bi+1.

In any case, we can find bn+1 as desired. So TDLO has quantifier elimination by Corollary 5.48.

Proposition 5.50 The theory TRG of the random graph has quantifier elimination.

226 First-order theories

Proof Since TRG has a finite relational vocabulary, we can apply Corollary 5.48. Let U be the set of vertices of the random graph GR. Let (a1, . . . , an) and (b1, . . . , bn) be two n-tuples of U that satisfy the same atomic formulas in GR. We must show that for any x U , there exists y U such that (a1, . . . , an, x) and (b1, . . . , bn, y) satisfy the same atomic formulas in GR. This follows from the fact that GR models the VR-sentence ρm for arbitrarily large m.

Example 5.51 Let V+ = {<, Ps, Pb} be an expansion of T< that includes two

unary relations Ps and Pb. Let TDLOE+ be the expansion of TDLOE to a V+- theory that interprets Ps(x) as the smallest element and Pb(x) as the biggest

element in the order. Then this theory has quantifier elimination. Recall from Example 5.47 that TDLOE does not have quantifier elimination. It was shown in that example that y(x < y) is not TDLOE -equivalent to a quantifier-free formula. This formula is TDLOE+ -equivalent to the quantifier-free formula ¬Pb(x). To show that TDLOE+ has quantifier elimination, we can use an argument similar to the proof of Proposition 5.49.

Example 5.52 Let T be an VE -theory that says E is an equivalence relation having infinitely many infinite classes. We show that T has quantifier elimination. Let M be the VE -structure having denumerably many equivalence classes each of which is denumerable. Let (a1, . . . , an) and (b1, . . . , bn) be two n-tuples of elements from the universe UM of M that satisfy the same atomic VE -formulas in M . Let an+1 be any element of UM . Since there are infinitely many equivalence classes in UM and each is infinite, we can surely find bn+1 UM such that M |= E(bi, bn+1) if and only if M |= E(ai, an+1).

So T has quantifier elimination. Now suppose that we expand the vocabulary by adding a unary relation P . Let TP be the expansion of T to the vocabulary {<, P } that says that P (x) holds for exactly one element x. Let N1 |= TP . Let a be the unique element such that N1 |= P (a). Let b be an element that is equivalent, but not equal, to a. Let c be an element that is not equivalent to a. Then b and c satisfy the same atomic formulas in N1 (since ¬P (b) and ¬P (c) both hold). Since there is no element y so that (b, a) and (c, y) satisfy the same atomic formulas, TP does not have quantifier elimination. In particular, the formula y(P (y) E(x, y)) is not TP -equivalent to a quantifier-free formula.

Now expand the vocabulary again to include the constant u. Let TP (u) be the expansion of TP to the vocabulary {<, P , u} that interprets u as the unique element for which P (u) holds. Since we have merely provided a name for an element that was already uniquely defined, TP (u) is essentially the same as TP (the models of TP can easily be viewed as models of TP (u) and vice versa). However, in contrast to TP , TP (u) does have quantifier elimination. The formulay(P (y) E(x, y)) is TP (u)-equivalent to the atomic formula E(x, u).

The previous examples demonstrate that quantifier elimination is a purely syntactic property. The theories TDLOE and TDLOE+ (like the theories TP and TP (u)) have very similar models. Given any model M of one of these theories, we can find a model N of the other theory so that M and N have the same underlying set and the same definable subsets.

Definition 5.53 Let M1 be a V1-structure and M2 be a V2-structure having the same underlying set. If the V1-definable subsets of M1 are the same as the V2-definable subsets of M2 then M1 and M2 are said to be bi-definable.

Two theories T1 and T2 are bi-definable if every model of T1 is bi-definable with some model of T2 and vice versa.

The theories TDLOE and TDLOE+ are bi-definable as are TP and TP (u). However, TDLOE+ and TP (u) have quantifier elimination whereas TDLOE and TP do not. The following proposition states that any theory has a bi-definable theory

with quantifier elimination.

Proposition 5.54 Let T be a V-theory. There exists a theory Tm that is bidefinable with T and has quantifier elimination.

Proof For each V-formula ϕ(x1, . . . , xn) (with n N), let Rϕ be an n-ary relation that is not in V. Let Vm = V {Rϕ|ϕ is a V-formula}. Let Tm be the expansion of T to a Vm-theory that contains the sentence ϕ(¯x) ↔ Rϕ(¯x) for each V-formula ϕ. Since each relation Rϕ is explicitly defined by Tm in terms of V, T , and Tm are bi-definable. Since every Vm-formula is Tm-equivalent to a quantifier-free formula, Tm has quantifier elimination.

The theory Tm in the previous proof is called the Morleyization of T . Morleyizations demonstrate that the property of quantifier elimination is not always useful. To analyze a structure M , we should choose an appropriate vocabulary. Ideally, we want to find a vocabulary V so that M is bi-definable with some V -structure M where T h(M ) has quantifier elimination. We must also require that the atomic V -formulas (and the relations between these formulas) are readily understood. The Morleyization of Tm is often of no use in this regard. If the atomic V -formulas are too complicated, then the quantifier elimination of T h(M ) does not lend insight into the structure M .

We have restricted our attention in this section to examples of theories that are particularly nice. With the exception of TS from Example 5.46, each theory we have considered is bi-definable with a theory in a finite relational vocabulary that has quantifier elimination. This is a severe restriction. The astute reader may have anticipated the following fact.

Proposition 5.55 Let T be a complete theory in a finite relational vocabulary. If T has quantifier elimination, then T is 0-categorical.

The elements a0 and b0 satisfy the same atomic formulas in M (as does each element of UM ). However, there is no element y so that (a0, y) and (b0, c0) satisfy the same atomic formulas.

If the vocabulary of T were finite and relational, then we could conclude that it does not have quantifier elimination. However, the vocabulary of T contains the function s. In fact, T is both complete and has quantifier elimination.

Example 5.57 Let V = {E, Pi|i N} be the vocabulary consisting of a binary relation E and denumerably many unary relations Pi. Let M be a countable V-structure that interprets E as an equivalence relation that has infinitely many equivalence classes of size 1, infinitely many equivalence classes of size 2, and no other equivalence classes. Moreover, each Pi holds for exactly two elements that are in the same equivalence class. So if a is in a class of size 1, then ¬Pi(a) holds for each i N. To complete our description of M , if M |= E(b, c) ¬(b = c) then M |= Pi(b) Pi(c) for exactly one i.

Let UM be the underlying set of M . If two tuples of elements from UM satisfy the same atomic V-formulas in M , then they satisfy the same V-formulas in M . However, T h(M ) does not have quantifier elimination. The formulay(¬(x = y) E(x, y)) is not T h(M )-equivalent to a quantifier-free formula.

Proposition 5.45 provides a necessary and su cient condition for a complete theory T in a finite relational vocabulary to have quantifier elimination. We restate this condition.

(ii) For any model M of T and any n N, if (a1, . . . , an) and (b1, . . . , bn) are n-tuples of UM that satisfy the same atomic formulas in M , then for any an+1 UM there exists bn+1 UM such that (a1, . . . , an, an+1) and (b1, . . . , bn, bn+1) satisfy the same atomic formulas in M (where UM denotes the underlying set of M ).

Corollary 5.48 states that, if the vocabulary V of T is finite and relational, then T has quantifier elimination if and only if condition (ii) holds for some model M of T . If V is not finite and relational, then, as Example 5.57 demonstrates, we must verify (ii) for more than one model. So Corollary 5.48 fails for vocabularies that are not finite and relational. Example 5.56 demonstrates that one direction of Proposition 5.45 also fails for vocabularies that are not finite and relational. Condition (ii) does not necessarily hold for all theories that have quantifier elimination.

It is still true that (ii) implies quantifier elimination. The proof that (ii) implies (i) in Proposition 5.45 makes no use of the hypothesis that the vocabulary is finite and relational. So (ii) is a su cient condition for quantifier elimination,