我正在尝试证明以下函数是真的,并且我很难搞清楚它,即使它看起来很明显!
(implies (and (listp x)
(listp y))
(equal (app (rev x) (rev y))
(rev (app x y))))
通过这样做我只需要使用函数显示(app(rev x)(rev y))等同于(rev(app x y)))):
(defunc len (x)
:input-contract t
:output-contract (natp (len x))
(if (atom x)
0
(+ 1 (len (rest x)))))
(defunc atom (x)
:input-contract t
:output-contract (booleanp (atom x))
(not (consp a)))
(defunc not (a)
:input-contract (booleanp a)
:output-contract (booleanp (not a))
(if a nil t))
(defunc listp (l)
:input-contract t
:output-contract (booleanp (listp l))
(or (equal l ())
(consp l)))
(defunc endp (a)
:input-contract (listp a)
:output-contract (booleanp (endp a))
(not (consp a)))
(defunc twice (l)
:input-contract (listp l)
:output-contract (listp (twice l))
(if (endp l)
nil
(cons (first l) (cons (first l) (twice (rest l))))))
(defunc app (a b)
:input-contract (and (listp a) (listp b))
:output-contract (listp (app a b))
(if (endp a)
b
(cons (first a) (app (rest a) b))))
(defunc rev (x)
:input-contract (listp x)
:output-contract (and (listp (rev x)) (equal (len x) (len (rev x))))
(if (endp x)
nil
(app (rev (rest x)) (list (first x)))))
这是我如何做另一个(希望正确)
(implies (listp y)
(equal (len (rev (cons x y)))
(+ 1 (len (rev y)))))
“反向追加事物”
(rev (app (rev y) (rex x))) = (app x y)
(len (rev (cons x y))
= def of rev
len (app (rev y) (list x))
= rev的输出合约
(len (rev (app (rev y) (list x))))
=“反向追加事物”
(len (rev (cons x y))
= rev的输出合约
(len (cons x y))
= def of len
(+ 1 (len y))
= rev的输出合约
(+ 1 (len (rev y)))