Major Section: FREE-VARIABLES-EXAMPLES

The following examples illustrate ACL2's handling of free variables in rewrite rules, as well as user control over how such free variables are handled. See free-variables for a background discussion.

(defstub p2 (x y) t) ; introduce unconstrained function; Get warning because of free variable. This would be an error if you had ; first executed (set-match-free-error t) in order to force yourself to ; specify :match-free (illustrated later, below). (defaxiom p2-trans (implies (and (p2 x y) (p2 y z)) (p2 x z)))

; Succeeds. (thm (implies (and (p2 a c) (p2 a b) (p2 c d)) (p2 a d)))

; The following causes an error because p2-trans is not a rune. (add-match-free-override :once p2-trans)

; After the following, the rewrite rule p2-trans will only allow one ; attempt per hypothesis to bind free variables. (add-match-free-override :once (:rewrite p2-trans))

; Now this same theorem fails to be proved. Here's why. The ; context for proving (p2 a d) happens to include the hypotheses in ; reverse order. So when the first hypothesis of p2-trans, namely ; (p2 x y), is relieved, where x is bound to a (as we are attempting ; to rewrite the current literal (p2 a d)), we find (p2 a b) in the ; context before (p2 a c) and hence y is bound to b. The ; instantiated second hypothesis of p2-trans is thus (p2 b d), and ; the proof fails. Before the add-match-free-override form above, ; the proof succeeded because the rewriter was allowed to backtrack ; and find the other binding for the first hypothesis of p2-trans, ; namely, y bound to c. Then the instantiated second hypothesis of ; p2-trans is (p2 c d), which is known to be true in the current ; context. (thm (implies (and (p2 a c) (p2 a b) (p2 c d)) (p2 a d)))

; Return to original behavior for binding free variables. (add-match-free-override :all t)

; Succeeds once again. (thm (implies (and (p2 a c) (p2 a b) (p2 c d)) (p2 a d)))

(u) ; undo (add-match-free-override :all t)

; This is an error, since no further arguments should appear after ; :clear. (add-match-free-override :clear t)

; Return all rules to original behavior for binding free variables, ; regardless of which previous add-match-free-override forms have ; been executed. (add-match-free-override :clear)

; This succeeds just as it did originally. (thm (implies (and (p2 a c) (p2 a b) (p2 c d)) (p2 a d)))

(ubt! 'p2-trans) ; back to the start, except retain the defstub

; Require that :match-free be specified for :linear and :rewrite rules with ; free variables. (set-match-free-error t)

; Fails because :match-free is missing. (defaxiom p2-trans (implies (and (p2 x y) (p2 y z)) (p2 x z)))

; Fails because :match-free must be followed by :once or :all. (defaxiom p2-trans (implies (and (p2 x y) (p2 y z)) (p2 x z)) :rule-classes ((:rewrite :match-free nil)))

; Succeeds, this time with no warning at all. (defaxiom p2-trans (implies (and (p2 x y) (p2 y z)) (p2 x z)) :rule-classes ((:rewrite :match-free :once)))

; Fails because we only bind once (see earlier long comment). (thm (implies (and (p2 a c) (p2 a b) (p2 c d)) (p2 a d)))

; Treat p2-trans as though `:match-free :all' had been specified. (add-match-free-override :all (:rewrite p2-trans))

; Succeeds since more than one binding is allowed for p2-trans. (thm (implies (and (p2 a c) (p2 a b) (p2 c d)) (p2 a d)))

(u) (u)

; Specify that future :linear and :rewrite rules with free variables ; that do not have :match-free specified are treated as though ; `:match-free :once' were specified. (set-match-free-default :once)

; Succeeds without error since `:match-free' is specified, as described ; above. But there is a warning, since :match-free is not specified for this ; :rewrite rule. (defaxiom p2-trans (implies (and (p2 x y) (p2 y z)) (p2 x z)))

; Fails since only single bindings are allowed for p2-trans. (thm (implies (and (p2 a c) (p2 a b) (p2 c d)) (p2 a d)))

; Treat p2-trans as though `:match-free :all' had been specified. (add-match-free-override :all t)

; Succeeds. (thm (implies (and (p2 a c) (p2 a b) (p2 c d)) (p2 a d)))

;;; Test searching of ground units, i.e. rewrite rules without ;;; variables on the left side of the conclusion, for use in ;;; relieving hypotheses with free variables. This is a very ;;; contrived example.

(ubt! 1) ; back to the start

(encapsulate (((p1 *) => *) ((p2 * *) => *) ((p3 *) => *) ((a) => *) ((b) => *)) (local (defun p1 (x) x)) (local (defun p2 (x y) (list x y))) (local (defun p3 (x) x)) (local (defun a () 0)) (local (defun b () 0)))

; Allow default of :match-free :all (form may be omitted). (set-match-free-error nil)

(defaxiom ax1 (implies (and (p2 x y) (p1 y)) (p3 x)))

(defaxiom p2-a-b (p2 (a) (b)))

(defaxiom p2-a-a (p2 (a) (a)))

(defaxiom p1-b (p1 (b)))

; Succeeds; see long comment below on next attempt to prove this ; theorem. (thm (implies (p2 (a) y) (p3 (a))))

; Now ax1 will only relieve hypothesis (p2 x y) for one binding of y: (add-match-free-override :once t)

; Fails when ax1 attempts to rewrite the conclusion to true, because ; the most recent ground unit for hypothesis (p2 x y) with x bound ; to (a) is rule p2-a-a, which binds y to (a). If more than one ground ; unit could be used then we would backtrack and apply rule p2-a-b, ; which binds y to (b) and hence hypothesis (p1 y) of ax1 is ; relieved by rule p1-b. (thm (implies (p2 (a) y) (p3 (a))))

; Return rules to original :match-free behavior. (add-match-free-override :clear)

; Succeeds once again. (thm (implies (p2 (a) y) (p3 (a))))

; Just for kicks, change the behavior of a built-in rule irrelevant ; to the proof at hand. (add-match-free-override :once (:rewrite string<-l-trichotomy))

; Still succeeds. (thm (implies (p2 (a) y) (p3 (a))))

;;;;;;;;;;

FINALLY, here is an example illustrating the use of the break-rewrite facility to get information about handling of free variables by the rewriter. Explanation is given after this (edited) transcript. Input begins on lines with a prompt (search for ``ACL2''); the rest is output.

ACL2 !>(encapsulate ((p1 (u x) t) (bad (x) t) (p2 (x y z) t) (bar (x y) t) (foo (x y) t) (poo (x y) t) (prop (u) t))The(local (defun p1 (u x) (declare (ignore u x)) nil)) (local (defun bad (x) (declare (ignore x)) nil)) (local (defun p2 (x y z) (declare (ignore x y z)) nil)) (local (defun bar (x y) (declare (ignore x y)) nil)) (local (defun foo (x y) (declare (ignore x y)) nil)) (local (defun poo (x y) (declare (ignore x y)) nil)) (local (defun prop (u) (declare (ignore u)) t))

(defthm foo-poo (implies (syntaxp (equal y 'y3)) (equal (foo x y) (poo x y))))

(defthm lemma-1 (implies (and (p1 u x) (bad x) (p2 x y z) (bar x y) (equal x x) ; admittedly silly! (foo x y)) (prop u)) :rule-classes ((:rewrite :match-free :all))))

; [[ output omitted ]]

Summary Form: ( ENCAPSULATE ((P1 ...) ...) ...) Rules: NIL Warnings: Subsume and Non-rec Time: 0.08 seconds (prove: 0.00, print: 0.01, other: 0.06) T ACL2 !>:brr t The monitored runes are: NIL T ACL2 !>:monitor (:rewrite lemma-1) t (((:REWRITE LEMMA-1) 'T)) ACL2 !>(thm (implies (and (p1 u0 x1) (bad x1) (bad x3) (bar x3 y1) (bar x3 y3) (p1 u0 x2) (p1 u0 x3) (p2 x3 y1 z1) (p2 x3 y3 z1)) (prop u0)))

(1 Breaking (:REWRITE LEMMA-1) on (PROP U0): 1 ACL2 >:eval

1x (:REWRITE LEMMA-1) failed because :HYP 1 contains free variables. The following display summarizes the attempts to relieve hypotheses by binding free variables; see :DOC free-variables.

[1] X : X1 Failed because :HYP 3 contains free variables Y and Z, for which no suitable bindings were found. [1] X : X2 Failed because :HYP 2 rewrote to (BAD X2). [1] X : X3 [3] Z : Z1 Y : Y1 Failed because :HYP 6 rewrote to (FOO X3 Y1). [3] Z : Z1 Y : Y3 Failed because :HYP 6 rewrote to (POO X3 Y3).

1 ACL2 >:unify-subst U : U0 1 ACL2 >

`:eval`

command above asks the rewriter to attempt to apply the rewrite
rule `lemma-1`

to the term `(prop u0)`

, shown just above the line with
`:eval`

. As we can see at the end, the variable `u`

in the conclusion of
`lemma-1`

is being bound to the variable `u0`

in the conjecture. The
first hypothesis of `lemma-1`

is `(p1 u x)`

, so the rewriter looks for
some `x`

for which `(p1 u0 x)`

is known to be true. It finds `x1`

, and
then goes on to consider the second hypothesis, `(bad x)`

. Since the
theorem we are proving has `(bad x1)`

in the hypothesis and `x`

is
currently bound to `x1`

, the rewriter is satisfied and moves on to the
third hypothesis of `lemma-1`

, `(p2 x y z)`

. However, `x`

is bound
to `x1`

and there are no instances of `y`

and `z`

for which
`(p2 x1 y z)`

is known in the current context. All of the above analysis
is summarized in the first part of the output from `:eval`

above:
[1] X : X1 Failed because :HYP 3 contains free variables Y and Z, for which no suitable bindings were found.Thus, the binding of

`x`

to `x1`

on behalf of the first hypothesis has
failed.
The rewriter now backs up to look for other values of `x`

that satisfy the
first hypothesis, and finds `x2`

because our current theorem has a
hypothesis of `(p1 u0 x2)`

. But this time, the second hypothesis of
`lemma-1`

, `(bad x)`

, is not known to be true for `x`

; that is,
`(bad x2)`

does not rewrite to `t`

; in fact, it rewrites to itself. That
explains the next part of the output from `:eval`

above:

[1] X : X2 Failed because :HYP 2 rewrote to (BAD X2).

The rewriter now backs up again to look for other values of `x`

that
satisfy the first hypothesis, and finds `x3`

because our current theorem
has a hypothesis of `(p1 u0 x3)`

. This time, the second hypothesis of
`lemma-1`

is not a problem, and moreover, the rewriter is able to bind
`y`

and `z`

to `y1`

and `z1`

, respectively, in order to satisfy the
third hypothesis, `(p2 x y z)`

: that is, `(p2 x2 y1 z1)`

is known in the
current context. That explains more of the above output from `:eval`

:

[1] X : X3 [3] Z : Z1 Y : Y1Unfortunately, the sixth hypothesis,

`(foo x y)`

, rewrites to itself
under the above bindings:
Failed because :HYP 6 rewrote to (FOO X3 Y1).So the rewriter looks for other bindings to satisfy the third hypothesis and finds these.

[3] Z : Z1 Y : Y3This time, the sixth hypothesis can be rewritten under the above bindings, from

`(foo x3 y3)`

to `(poo x3 y3)`

by lemma `foo-poo`

, but still not
to `t`

.
Failed because :HYP 6 rewrote to (POO X3 Y3).There are no more free variable bindings to try, so this concludes the output from

`:eval`

.