CONSERVATIVITY-OF-DEFCHOOSE

proof of conservativity of defchoose
Major Section:  DEFCHOOSE


This documentation topic provides underlying theory. It is of theoretical interest only; it has no relationship to the effective use of ACL2.

The argument below for the conservativity of defchoose replaces the terse and somewhat misleading reference to a forcing argument in Appendix B of the paper by ACL2 authors Kaufmann and Moore, Structured Theory Development for a Mechanized Logic'' (Journal of Automated Reasoning 26, no. 2 (2001), pp. 161-203).

Our basic idea is to to take a (countable) first-order structure for ACL2, M, together with a function symbol, f, introduced by defchoose, and find a way to expand M with an interpretation of f (without changing the universe of M) so that e0-induction continues to hold in the expansion. A remark at the end of this documentation topic shows why care is necessary. A concept called forcing'', originally introduced by Paul Cohen for set theory, has long since been adapted by logicians (in a simplified form) to model theory. This simplified model-theoretic forcing provides the means for making our careful expansion.

The forcing argument presented below is intended to be completely self-contained for those familiar with basic first-order logic and ACL2. No background in forcing (model-theoretic or otherwise) is expected, though we do expect a rudimentary background in first-order logic and familiarity with the following.

Preliminaries. We write s[p<-p0] to denote the result of extending or modifying the assignment s by binding p to p0. Now let A be a subset of the universe U of a first-order structure M. A is said to be first-order definable with parameters'' in M if for some formula phi, variable x, and assignment s binding the free variables of phi except perhaps for x, A = {a \in U: M |= phi[s[x<-a]]. Note that we are writing \in'' to denote set membership. Finally, we indicate the end of a proof (or of a theorem statement, when the proof is omitted) with the symbol -|''.

We gratefully acknowledge very helpful feedback from John Cowles, who found several errors in a draft of this note and suggested the exercises. We also thank Ruben Gamboa for helpful feedback, and we thank Jim Schmerl for an observation that led us directly to this proof in the first place.

We are given a consistent first-order theory T, extending the ACL2 ground-zero theory, that satisfies the e0-induction scheme. We wish to show that the extension of T by the following arbitrary defchoose event is conservative, where g is a new function symbol.

     (defchoose g <bound-vars> <free-vars> <body>)

Note that by the extension of T'' here we mean the extension of T by not only the new defchoose axiom displayed just below, but also the addition of e0-induction axioms for formulas in the language with the new defchoose function symbol, g.
     <body> -> (LET <free-vars> = g(<bound-vars>) in <body>)

By definition of conservativity, since proofs are finite, it clearly suffices to consider an arbitrary finite subset of T. Then by the completeness, soundness, and downward Lowenheim-Skolem theorems of first-order logic, it suffices to show that an arbitrary countable model of T can be expanded (i.e., by interpreting the new symbol g without changing the universe of the model) to a model of the corresponding defchoose axiom above, in which all e0-induction axioms hold in the language of that model.

Below, we will carry out a so-called forcing construction, which allows us to expand any countable model M of T to a model M[G] that satisfies e0-induction and also satisfies the above axiom generated from the above defchoose event. The ideas in this argument are standard in model theory; no novelty is claimed here.

Fix a countable model M of a theory T that satisfies e0-induction and extends the ACL2 ground-zero theory. Also fix the above defchoose axiom, where g is not in the language of T.

We start by defining a partial order P as follows. Let Nb and Nf be the lengths of <bound-vars> and <free-vars>, respectively. P consists of all fn in M such that the following formula is true in M. Roughly speaking, it says that fn is a finite function witnessing the above requirement for g.

       alistp(fn) &
no-duplicatesp-equal(strip-cars(fn)) &
(forall <bound-vars>, <free-vars> .
(member-equal(cons(<bound-vars>,<free-vars>), fn) ->
(length(<bound-vars>) = Nb &
length(<free-vars>)  = Nf &
((exists <free-vars> . <body>) -> <body>))))

P is ordered by subset, i.e., we say that p2 extends p1 if p1 is a subset (not necessarily proper) of p2 (more precisely, M |= subsetp-equal(p1,p2)).

Remark. The original argument in Appendix B of the aforementioned paper can essentially be salvaged, as we now show. The key observation is that the particular choice of P is nearly irrelevant for the argument that follows below. In particular, we can instead define P to consist of finite one-one functions with domain contained in the set of natural numbers. More precisely, consider the following definitions.

     (defun function-p (fn)
(declare (xargs :guard t))
(and (alistp fn)
(no-duplicatesp-equal (strip-cars fn))))

(defun nat-listp (x)
(declare (xargs :guard t))
(cond ((atom x)
(equal x nil))
(t (and (natp (car x))
(nat-listp (cdr x))))))

(defun nat-function-p (x)
(and (function-p x)
(nat-listp (strip-cars x))))

and define inverse as follows.
     (defun inverse (fn)
(declare (xargs :guard (alistp fn)))
(if (endp fn)
nil
(cons (cons (cdar fn) (caar fn))
(inverse (cdr fn)))))

Then P may instead be defined to consist of those fn for which nat-function-p(fn) & function-p(inverse(fn)). With this alternate definition of P, the argument below then goes through virtually unchanged, and we get an expansion M[G] of M in which there is a definable enumeration of the universe. The conservativity of defchoose then follows easily because the function being introduced can be defined explicitly using that enumeration (namely, always pick the least witness in the sense of the enumeration).

End of Remark.

Next we present the relevant forcing concepts from model theory.

A dense subset of P is a subset D of P such that for every p \in P, there is d \in D such that d extends p. A subset G of P is generic with respect to a collection Ds of dense subsets of P, also written G is Ds-generic,'' if G is closed under subset (if p2 \in G and p2 extends p1 then p1 \in G), G is pairwise compatible (the union-equal of any two elements of G is in G), and every set in Ds has non-empty intersection with G.

For p \in P, we say that a subset D of P is dense beyond p if for all p1 extending p there exists p2 extending p1 such that p2 \in D. This notion makes sense even for D not a subset of P if we treat elements of D not in P as nil.

Proposition 1. For any partial order P and countable collection Ds of dense subsets of P, there is a Ds-generic subset of P.

Proof. Let Ds = {D0,D1,D2,...}. Define a sequence <p_0,p_1,...> such that for all i, p_i \in Di and p_(i+1) extends p_i. Let G = {p \in P: for some i, pi extends p}. Then G is Ds-generic. -|

Note that P is first-order definable (with parameters) in M. Let Df be the set of dense subsets of P that are first-order definable (with parameters) in M. A standard argument shows there are only countably many first-order definitions with parameters in a countable model M -- for example, we can Goedel number all terms and then all formulas -- hence, Df is countable.

By Proposition 1, let G be Df-generic. Notice that for any list x of length Nb in M, the set of elements f of P for which x is in the domain of f is dense and first-order definable. We may thus define a function g0 as follows: g0(x_1,...,x_Nb) = y if there is some element of G containing the pair ((x_1 ... x_Nb) . y). It is easy to see that g0 is a total function on M. Let L be the language of T and let L[g] be the union of L with a set containing a single new function symbol, g. Let M[G] be the expansion of M to L[g] obtained by interpreting g to be g0 (see also Proposition 5 below).

So now we have fixed M, P, Df, G, and g0, where G is Df-generic.

Proposition 2. Let Df be the set of dense subsets of P that are first-order definable (with parameters) in M. Suppose that p \in G and D \in Df. Then for some q \in G extending p, q \in D.

Proof. Let D0 be the set of p' \in D that either extend p or have no extension in D that extends p. We leave it as a straightforward exercise to show that D0 is dense, and D0 is clearly first-order definable (with parameters) in M. So by genericity of G, we may pick q \in D0 such that q \in G. Thus q \in D. By definition of generic, some extension q1 of both p and q belongs to G. Pick q2 \in D extending q1; thus q has an extension in D that extends p (namely, q2), so by definition of D0, q extends p. -|

Definition of forcing. Let phi(x1,...,xk) be a first-order formula in L[g] and let p \in P. We define a formula of L, denoted p ||- phi'' (p forces phi''), by recursion on phi (in the metatheory) as follows. (Here, we view or'' and forall'' as abbreviations.)

If phi is atomic, then let phi'(A) be the result of replacing, inside-out, each subterm of the form g(x_1,...,x_Nb) with the term (cdr (assoc-equal (list x_1 ... x_Nb) A)), where A is neither p nor a variable occurring in phi. Then p ||- phi is defined as follows: The set {A \in P: A extends p and phi'(A)} is dense beyond p''. That is, p ||- phi is the following formula:
    (forall p1 \in P extending p)
(exists p2 \in P extending p1) phi'(p2).

p ||- ~phi is: (forall p' \in P extending p) ~(p' ||- phi)

p ||- phi_1 & phi_2 is: (p ||- phi_1) & (p ||- phi_2)

p ||- (exists x) phi is: (exists x) (p ||- phi)

We will need the following definition later.

Definition. p ||-w phi (p weakly forces phi) is an abbreviation for p ||- ~~phi.

The following exercises were suggested by John Cowles as a means for gaining familiarity with the definition of forcing.

Exercise 1. Consider the formula (phi_1 OR phi_2) as an abbreviation for ~(~phi_1 & ~phi_2), Show that p ||- (phi_1 OR phi_2) is equivalent to the following.

     (forall p' \in P extending p) (exists p'' \in P extending p')
((p'' ||- phi_1) OR (p'' ||- phi_2))


Exercise 2. Consider the formula (forall x)phi as an abbreviation for ~(exists x)~phi, Show that p ||- (forall x)phi is equivalent to the following.

     (forall x)
(forall p1 \in P extending p)
(exists p2 \in P extending p1) (p2 ||- phi).


Exercise 3. Prove that p ||-w phi is equivalent to the following.

     (forall p' \in P extending p)
(exists p'' \in P extending p') (p'' ||- phi).


Exercise 4. Let phi be a formula of L[g]. Prove: M |= (p ||- phi)[s[p<-p0]] implies M |= (p ||-w phi)[s[p<-p0]].

Exercise 5. Let phi be a formula of L[g]. Prove: M |= (p ||- ~phi)[s[p<-p0]] iff M |= (p ||-w ~phi)[s[p<-p0]].

[End of exercises.]

The definition of forcing stipulates how to view p ||- phi(x1,...,xk)'' as a new formula theta(p,x1,...,xk). That is, ||-'' transforms formulas, so for any first-order formula phi, p ||- phi'' is just another first-order formula. That observation shows that a formula such as ((p ||- phi) OR (p ||- ~phi)) is really just another first-order formula. The following proposition thus follows easily.

Proposition 3. For any formula phi of L[g], {p0: M |= ((p ||- phi) OR (p ||- ~phi))[s[p<-p0]]]} is a dense subset of P, which (since it is first-order definable with parameters in M) intersects G. -|

The following proposition is easily proved by a structural induction on phi, and is left to the reader.

Proposition 4. Let phi be a formula of L[g]. Suppose p0 in P, p1 in P,
M |= (p ||- phi)[s[p<-p0]] and p1 extends p0. Then
M |= (p ||- phi)[s[p<-p1]]. -|

We will also need the following.

Proposition 5. The following is dense for any finite set S of Nb-tuples: {p \in P: for some <x_1 ... x_Nb> \in S, (list x_1 ... x_Nb) \in strip-cars(p)}. Thus, the function g0 is a total function. -|

The next lemma tells us that the sentences true in M[G] are those that are forced by an element of G.

Truth Lemma. Let phi be a formula in L[g], let s be an assignment to the free variables of phi, and let p be a variable not in the domain of s. Then M[G] |= phi[s] iff for some p0 \in G, M |= (p ||- phi)[s[p<-p0]].

Proof. The proof is by induction on the structure of phi. First suppose phi is atomic. Let D* be the set of elements p0 \in P such that every assoc-equal evaluation from the definition of forcing phi returns a pair when A is bound to p0. (Intuitively, this means that p0 is a sufficiently large approximation from any G containing p0 to make sense of phi in M[G].) We make the following claim.

(*)   For all p0 \in G such that p0 \in D*,
M[G] |= phi[s] iff M |= (p ||- phi)[s[p<-p0]].


To prove the claim, fix p0 in both G and D*, and recall the function g0 constructed from G in the definition of M[G]. Suppose that t_1, ..., t_Nb are terms and g(t_1, ..., t_Nb) is a subterm of phi. Then s assigns a value in M to each of the t_i. Let a_i be the value assigned by s to t_i. Then g0(a_1, ..., a_Nb) = (cdr (assoc-equal (list a_1 ... a_Nb) p0)), as the assoc-equal is a pair (since p0 \in D*) and has the indicated value (because p0 \in G). It follows by the definition of formula phi' in the definition of forcing:

     M[G] |= phi[s]  iff  M |= phi'(p)[s[p<-p0]]

Moreover, because p0 \in D* it is clear that this holds if p0 is replaced by an arbitrary extension of p0. Then (*) easily follows.

By Proposition 5, D* is dense, so there is some p0 in the intersection of D* and G. The forward direction of the conclusion then follows by (*). The reverse direction is clear from (*) by application of Proposition 2 to D* and Proposition 4.

Next, suppose M[G] |= ~phi[x]. Then it is not the case that M[G] |= phi, so by the inductive hypothesis, there is no p0 \in G for which M |= (p ||- phi)[s[p<-p0]]. By Proposition 3, there is p0 \in G for which M |= (p ||- ~phi)[s[p<-p0]]. For the other direction, suppose it is not the case that M[G] |= ~phi[s]. So M[G] |= phi[s], and by the inductive hypothesis, there is p0 \in G for which M |= (p ||- phi)[s[p<-p0]]. It follows that there is no p1 \in G for which M |= (p ||- ~phi)[s[p<-p1]], since from such p1 we can find a common extension p2 of p0 and p1 (since G is generic), and since p2 extends p0 then by Proposition 4, M |= (p ||- phi)[s[p<-p2]], contradicting (by definition of forcing) M |= (p ||- ~phi)[s[p<-p1]] since p2 extends p1.

The case (phi_1 & phi_2) follows easily from the inductive hypothesis. For the forward direction, apply Proposition 4 and the observation that by genericity, if p0 \in G and p1 \in G then p0 and p1 they have a common extension in G.

Finally, the case (exists x) phi follows trivially from the inductive hypothesis. -|

Truth Lemma Corollary. The Truth Lemma holds with ||-w replacing ||-.

Proof. This is clear by applying the Truth Lemma to ~~phi. -|

Here is our main theorem. Recall that all first-order theories in our ACL2 context satisfy the e0-induction scheme.

Theorem. M[G] satisfies e0-induction.

Proof. We consider an arbitrary instance of e0-induction in L[g], stated using a strict well-founded relation <| and a formula phi. We write phi(y) to indicate that y may be among the free variables of phi, and phi(y<-x) to denote the result of substituting x for y in phi.

  theta(y):   (forall y) [((forall x <| y) phi(y<-x)) -> phi(y)]
-> (forall y) phi(y)

Our goal is to prove that theta holds in M[G].

Below, we abuse notation by leaving assignments implicit and by writing p ||- phi(y0)'' to signify that the formula (p ||- phi(y)) is true in M under the extension of the explicit assignment that binds y to y0. We believe that the intended meaning will be clear.

Consider the following set D.

  D = {p \in P: either p ||-w phi(y0) for all y0,
or else
for some y0, p ||- ~phi(y0) and
for all y1 <| y0 p ||-w phi(y1)}.

The set D is clearly first-order definable (with parameters) in M. We claim that D is a dense subset of P. For suppose p0 \in P; we find p1 \in D extending p0, as follows. If p0 ||-w phi(y0) for all y0, then we may take p1 to be p0. Otherwise, by definition of ||-w and ||-, there is some y0 such that for some extension p0' of p0, p0' ||- ~phi(y0). Pick a <|-minimal such y0, and correspondingly pick p1 so that p1 extends p0 and p1 ||- ~phi(y0). In order to show that p1 \in D, it remains to show that for all y1 <| y0, p1 ||-w phi(y1), i.e., there is no q extending p1 such that q ||- ~phi(y1). This is indeed the case since otherwise q and y1 would contradict the <|-minimality of y0.

Applying the genericity of G and just-proved density of D, pick p0 \in G such that p0 \in D. If p0 ||-w phi(y0) for all y0, then by the Truth Lemma Corollary, M[G] |= phi(y0) for all y0, and thus M[G] |= theta. Otherwise, since p0 \in D we may choose y0 such that p0 ||- ~phi(y0) and for all y1 <| y0, p0 ||-w phi(y1). By the Truth Lemma and its corollary, since p0 \in G we have:

(1)   M[G] |= ~phi(y0).
(2)   For all y1 <| y0, M[G] |= phi(y1).

It follows that the antecedent of theta is false in M[G], as witnessed by y = y0; thus M[G] |= theta. -|

Remark. We close by returning, as promised above, to the question of why so much care is necessary in constructing an expansion of M. We assume familiarity here with the notion of a non-standard'' natural number of M, i.e., one that is greater than the interpretation of any term that has the form (+ 1 1 1 ... 1). Here is a very simple example that illustrates the need for some care. Consider the following event, which introduces a function foo with the following property: for all x, if natp(x) then natp(foo(x)).

     (defchoose foo (y) (x)
(implies (natp x) (natp y)))

Certainly we can build a model of the above property from a model M of the ground-zero theory, by interpreting foo so that for all x for which M satisfies natp(x), foo(x) is also a natp in M. But suppose we start with a non-standard model M of the ground-zero theory, and we happen to define foo(x) to be 1 for all non-standard natural numbers x and 0 for all other x. The resulting expansion of M will not satisfy the e0-induction scheme or even the ordinary natural number induction scheme: foo(0)=0 holds in that expansion as does the implication foo(n)=0 => foo(n+1)=0 for every natural number n of M, standard or not; and yet foo(k)=0 fails for every non-standard natural number k of M.