C Proof of Theorem 4 . 1 a Proof of Theorem 3 . 1
نویسنده
چکیده
26 2. 1 is not a proof. From the induction hypotheses we have that`MT 1 = fail. If s 0 is either an axiom or an assumption, then is apply(then(t 1 ; \t "); \s 0 ") and`MT Tac(\s 0 "). If s 0 is neither an axiom nor an assumption, then is apply(then(t 1 ; \t "); fail) with`MT Tac(fail). In both cases, from axiom (A9), the induction hypotheses and by applying ifE we havè MT = fail 25 Theorem C.1 Let be a sequent tree of s. Let be a sequential tactic application of. If is a proof of s, then`MT = \s" and`MT T(\s"). Proof : Base Case: If is s, then it must be either an axiom or an assumption. Then T(\s") is an axiom of MT. is apply(\idtac"; \s"). From axioms (A7) and (A5) we havè MT apply(\idtac"; \s ") = \s ". Step Case : is apply(then(t 1 ; \t ") ; \s 0 ") and 1 is apply(t 1 ; \s 0"), since and 1 are proofs. From the induction hypotheses`MT 1 = \s 1 " and`MT T (\s1 "). From axiom (A9) : ` MT = if (\s 1 " = fail) then fail else apply (\t " ; \s 1 ") (9) Sincè MT \s 1 " 6 = fail , we apply If E : and obtain`MT = apply (\t " ; \s 1 ")) (10) From axiom (A4) and (A3) , we havè MT = if (: Fail (\s 1 ") ^ P (\s 1 ")) then f (\s 1 ") else fail (11) We havè MT : Fail (\s 1 ") from axiom (A2) and`MT P (\s 1 ") and`MT f (\s 1 ") = \s " (since must be applicable to s 1). Therefore (rule if E) ` MT = \s ". From axiom (A1) we havè MT (T (\s 1 ") ^ P (\s 1 ")) T (f (\s 1 ")) (12) and then`MT T (f (\s 1 ")). Thereforè MT T (\s "). Theorem C. 2 Let be a sequent tree of s. Let be a sequential tactic application of. If is not a proof , then`MT = fail Proof : Base Case : If is s , then it is neither an axiom nor an assumption. Then is apply (\idtac " ; fail) that is provably equal to fail (axioms (A7) and (A5)). Step …
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تاریخ انتشار 1994