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snark-20120808r02/examples/front-last-example.lisp

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+;;; -*- Mode: Lisp; Syntax: Common-Lisp; Package: snark-user -*-
+;;; File: front-last-example.lisp
+;;; The contents of this file are subject to the Mozilla Public License
+;;; Version 1.1 (the "License"); you may not use this file except in
+;;; compliance with the License. You may obtain a copy of the License at
+;;; http://www.mozilla.org/MPL/
+;;;
+;;; Software distributed under the License is distributed on an "AS IS"
+;;; basis, WITHOUT WARRANTY OF ANY KIND, either express or implied. See the
+;;; License for the specific language governing rights and limitations
+;;; under the License.
+;;;
+;;; The Original Code is SNARK.
+;;; The Initial Developer of the Original Code is SRI International.
+;;; Portions created by the Initial Developer are Copyright (C) 1981-2002.
+;;; All Rights Reserved.
+;;;
+;;; Contributor(s): Mark E. Stickel <stickel@ai.sri.com>.
+
+(in-package :snark-user)
+
+;;; Let L be a nonempty list.
+;;; Synthesize a program to compute the FRONT and LAST
+;;; of the list where LAST of a list is its last element
+;;; and FRONT is the list of all elements except the last.
+;;;
+;;; The program specification is
+;;;   (EXISTS (Y Z) (= L (APPEND Y (CONS Z NIL))))
+;;; i.e., find Y and Z such that L can be formed by
+;;; appending Y (the FRONT of L) and a single element list
+;;; containing Z (the LAST of L).
+;;;
+;;; The appropriate inductive axiom is given explicitly in the axiom
+;;; named induction.
+;;;
+;;; Necessary properties of APPEND, CONS, HEAD, and TAIL are given
+;;; in the axioms named append-nil, append-cons, and cons-definition.
+;;;
+;;; A proof of the query is found and the program
+;;; defined by the values found for variables Y and Z
+;;; in the specification.
+;;;
+;;; Resolution and paramodulation (for equality) are the inference
+;;; rules used.
+
+(defun front-last-example ()
+  ;; Waldinger program synthesis example 1989-12-14
+  (initialize)
+  (use-resolution)
+  (use-paramodulation)
+  (use-literal-ordering-with-resolution     'literal-ordering-p)
+  (use-literal-ordering-with-paramodulation 'literal-ordering-p)
+  (use-conditional-answer-creation)
+  (declare-constant 'nil)
+  (declare-constant 'l)    
+  (declare-function 'head   1)
+  (declare-function 'tail   1)
+  (declare-function 'cons   2)
+  (declare-function 'append 2)
+  (declare-function 'front  1)
+  (declare-function 'last   1)
+  (declare-ordering-greaterp 'l 'nil)
+  (declare-ordering-greaterp 'head 'l)
+  (declare-ordering-greaterp 'tail 'l)
+  (declare-ordering-greaterp 'cons 'head)
+  (declare-ordering-greaterp 'cons 'tail)
+  (declare-ordering-greaterp 'append 'cons)
+;;(assert '(forall (x) (= x x)))
+  (assert '(/= l nil)
+	  :name 'l-nonempty)
+  (assert '(implies (and (/= l nil) (/= (tail l) nil))
+                    (= (tail l) (append (front (tail l)) (cons (last (tail l)) nil))))
+	  :name 'induction)
+  (assert '(forall (u) (= (append nil u) u))
+	  :name 'append-nil)
+  (assert '(forall (u v w) (= (append (cons u v) w) (cons u (append v w))))
+	  :name 'append-cons)
+  (assert '(forall (x) (implied-by (= x (cons (head x) (tail x))) (/= x nil)))
+	  :name 'cons-definition)
+  (prove '(= l (append ?y (cons ?z nil))) :answer '(values ?y ?z)))
+
+;;; front-last-example.lisp EOF