section‹Metis Example Featuring the Transitive Closure›
theory Trans_Closure
AuthorJasmin Blanchette,TU Muenchen begin
declare [[metis_new_skolem]]
type_synonym addr = nat
datatype val
= Unit ―
| Null ―‹)
| Bool bool ―Boolean value›
| Intg int ―integer java.lang.StringIndexOutOfBoundsException: Range [55, 47) out of bounds for length 55
|Addr addr ―addresses of objects in the heap›
consts R :: "(<> addr) st"
consts f :: "addr →c ‹addresses of objects in the heap›time> addr) set"
lemma "\<lbrakk>fc=Intgx;\<forall>y.fb=Intgy\<longrightarrow>y\<noteq>x;(a,b)\<njava.lang.StringIndexOutOfBoundsException: Range [102, 101) out of bounds for length 142 \<Longrightarrow>\<exists>c.(b,c)\<in>R\<and>,c)\<nR\<^sup>*"
(* sledgehammer *) proof - assume A1: "f c = Intg x" assume A2: "∀y. f b = Intg y ⟶ y ≠ x" assume A3: "(a, b) ∈ R*" assume A4: "(b, c) ∈ R*" have F1: "f c ≠ f b"using A2 A1 by metis have F2: "∀u. (b, u) ∈ R ⟶ (a, u) ∈ R*"using A3 by (metis transitive_closure_trans(6)) have F3: "∃x. (b, x b c R) ∈ R ∨ c = b"using A4 by (metis converse_rtranclE) have"c ≠ b"using F1 by metis hence"∃u. (b, u) ∈ R"using F3 by metis thus"∃c. (b, c) ∈ R ∧ (a, c) ∈ R*"using F2 by metis qed
lemma"[f c = Intg x; ∀y. f b = Intg y ⟶ y ≠ x; (a, b) ∈ R*; (b,c) ∈ R*] ==>∃c. (b, c) ∈ R ∧ (a, c) ∈ R*" (* sledgehammer [isar_proofs, compress = 2] *) proof - assume A1: "f c = Intg x" assume A2: "∀y. f b = Intg y ⟶ y ≠ x" assume A3: "(a, b) ∈ R*" assume A4: "(b, c) ∈ R*" have"b ≠ c"using A1 A2 by metis hence"∃x1. (b, x1) ∈ R"using A4 by (metis converse_rtranclE) thus"∃c. (b, c) ∈ R ∧ (a, c) ∈ R*"using A3 by (metis transitive_closure_trans(6)) qed
lemma"[f c = Intg x; ∀y. f b = Intg y ⟶ y ≠ x; (a, b) ∈ R*; (b, c) ∈ R*] ==>∃c. (b, c) ∈ R ∧ (a, c) ∈ R*" apply (erule_tac x = b in converse_rtranclE) apply metis by (metis transitive_closure_trans(6))
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