Z3 で与えられた一般的なトポロジーの定理を証明しようとしています。
次のZ3-SMT-LIBコードを使用して、そこにあるコードを翻訳しています
;; File : TOP001-2 : TPTP v6.0.0. Released v1.0.0.
;; Domain : Topology
;; Problem : Topology generated by a basis forms a topological space, part 1
(declare-sort S)
(declare-sort Q)
(declare-sort P)
(declare-fun elemcoll (S Q) Bool)
(declare-fun elemset (P S) Bool)
(declare-fun unionmemb (Q) S)
(declare-fun f1 (Q P) S)
(declare-fun f11 (Q S) P)
(declare-fun basis (S Q) Bool)
(declare-fun Subset (S S) Bool)
(declare-fun topbasis (Q) Q)
;; union of members axiom 1.
(assert (forall ((U P) (Vf Q)) (or (not (elemset U (unionmemb Vf)))
(elemset U (f1 Vf U) ) ) ))
;; union of members axiom 2.
(assert (forall ((U P) (Vf Q)) (or (not (elemset U (unionmemb Vf)))
(elemcoll (f1 Vf U) Vf ) ) ))
;; basis for topology, axiom 28
(assert (forall ((X S) (Vf Q)) (or (not (basis X Vf)) (= (unionmemb Vf) X ) ) ))
;; Topology generated by a basis, axiom 40.
(assert (forall ((Vf Q) (U S)) (or (elemcoll U (topbasis Vf))
(elemset (f11 Vf U) U)) ))
;; Set theory, axiom 7.
(assert (forall ((X S) (Y Q)) (or (not (elemcoll X Y)) (Subset X (unionmemb Y) ) ) ))
;; Set theory, axiom 8.
(assert (forall ((X S) (Y S) (U P)) (or (not (Subset X Y)) (not (elemset U X))
(elemset U Y) )))
;; Set theory, axiom 9.
(assert (forall ((X S)) (Subset X X ) ))
;; Set theory, axiom 10.
(assert (forall ((X S) (Y S) (Z S)) (or (not (= X Y)) (not (Subset Z X)) (Subset Z Y) ) ))
;; Set theory, axiom 11.
(assert (forall ((X S) (Y S) (Z S)) (or (not (= X Y)) (not (Subset X Z)) (Subset Y Z) ) ))
(check-sat)
(push)
(declare-fun cx () S)
(declare-fun f () Q)
(assert (basis cx f))
(assert (not (elemcoll cx (topbasis f))))
(check-sat)
(pop)
(push)
(assert (basis cx f))
(assert (elemcoll cx (topbasis f)))
(check-sat)
(pop)
対応する出力は
sat
sat
sat
この例をここでオンラインで実行してください
最初satは正しいです。しかし、2 番目satは間違っていますunsat。つまり、Z3 は定理とその否定が同時に真であると言っているのです。
この場合どうなるか教えてください。どうもありがとう。ではごきげんよう。