Major Section: QUANTIFIERS

See quantifiers-using-defun-sk for the context of this example.

(in-package "ACL2"); We prove that if every member A of each of two lists satisfies the ; predicate (P A), then this holds of their append; and, conversely.

; Here is a solution using explicit quantification.

(defstub p (x) t)

(defun-sk forall-p (x) (forall a (implies (member a x) (p a))))

; The defun-sk above introduces the following axioms. The idea is that ; (FORALL-P-WITNESS X) picks a counterexample to (forall-p x) if there is one.

#| (DEFUN FORALL-P (X) (LET ((A (FORALL-P-WITNESS X))) (IMPLIES (MEMBER A X) (P A))))

(DEFTHM FORALL-P-NECC (IMPLIES (NOT (IMPLIES (MEMBER A X) (P A))) (NOT (FORALL-P X))) :HINTS (("Goal" :USE FORALL-P-WITNESS))) |#

; The following lemma seems critical.

(defthm member-append (iff (member a (append x1 x2)) (or (member a x1) (member a x2))))

; The proof of forall-p-append seems to go out to lunch, so we break into ; directions as shown below.

(defthm forall-p-append-forward (implies (forall-p (append x1 x2)) (and (forall-p x1) (forall-p x2))) :hints (("Goal" ; ``should'' disable forall-p-necc, but no need :use ((:instance forall-p-necc (x (append x1 x2)) (a (forall-p-witness x1))) (:instance forall-p-necc (x (append x1 x2)) (a (forall-p-witness x2)))))))

(defthm forall-p-append-reverse (implies (and (forall-p x1) (forall-p x2)) (forall-p (append x1 x2))) :hints (("Goal" :use ((:instance forall-p-necc (x x1) (a (forall-p-witness (append x1 x2)))) (:instance forall-p-necc (x x2) (a (forall-p-witness (append x1 x2))))))))

(defthm forall-p-append (equal (forall-p (append x1 x2)) (and (forall-p x1) (forall-p x2))) :hints (("Goal" :use (forall-p-append-forward forall-p-append-reverse))))