QUANTIFIERS-USING-DEFUN-SK

Major Section:  QUANTIFIERS

See quantifiers for the context of this example. It should be compared to a corresponding example in which a simpler proof is attained by using recursion in place of explicit quantification; see quantifiers-using-recursion.

(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))))


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


(defthm forall-p-append
  (equal (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)))
            (: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))))))))

Also see quantifiers-using-defun-sk-extended for an elaboration on this example.