--- /dev/null Thu Jan 01 00:00:00 1970 +0000
+++ b/src/HOL/Metis_Examples/Trans_Closure.thy Mon Jun 06 20:36:35 2011 +0200
@@ -0,0 +1,66 @@
+(* Title: HOL/Metis_Examples/Trans_Closure.thy
+ Author: Lawrence C. Paulson, Cambridge University Computer Laboratory
+ Author: Jasmin Blanchette, TU Muenchen
+
+Metis example featuring the transitive closure.
+*)
+
+header {* Metis Example Featuring the Transitive Closure *}
+
+theory Trans_Closure
+imports Main
+begin
+
+declare [[metis_new_skolemizer]]
+
+type_synonym addr = nat
+
+datatype val
+ = Unit -- "dummy result value of void expressions"
+ | Null -- "null reference"
+ | Bool bool -- "Boolean value"
+ | Intg int -- "integer value"
+ | Addr addr -- "addresses of objects in the heap"
+
+consts R :: "(addr \<times> addr) set"
+
+consts f :: "addr \<Rightarrow> val"
+
+lemma "\<lbrakk>f c = Intg x; \<forall>y. f b = Intg y \<longrightarrow> y \<noteq> x; (a, b) \<in> R\<^sup>*; (b, c) \<in> R\<^sup>*\<rbrakk>
+ \<Longrightarrow> \<exists>c. (b, c) \<in> R \<and> (a, c) \<in> R\<^sup>*"
+(* sledgehammer *)
+proof -
+ assume A1: "f c = Intg x"
+ assume A2: "\<forall>y. f b = Intg y \<longrightarrow> y \<noteq> x"
+ assume A3: "(a, b) \<in> R\<^sup>*"
+ assume A4: "(b, c) \<in> R\<^sup>*"
+ have F1: "f c \<noteq> f b" using A2 A1 by metis
+ have F2: "\<forall>u. (b, u) \<in> R \<longrightarrow> (a, u) \<in> R\<^sup>*" using A3 by (metis transitive_closure_trans(6))
+ have F3: "\<exists>x. (b, x b c R) \<in> R \<or> c = b" using A4 by (metis converse_rtranclE)
+ have "c \<noteq> b" using F1 by metis
+ hence "\<exists>u. (b, u) \<in> R" using F3 by metis
+ thus "\<exists>c. (b, c) \<in> R \<and> (a, c) \<in> R\<^sup>*" using F2 by metis
+qed
+
+lemma "\<lbrakk>f c = Intg x; \<forall>y. f b = Intg y \<longrightarrow> y \<noteq> x; (a, b) \<in> R\<^sup>*; (b,c) \<in> R\<^sup>*\<rbrakk>
+ \<Longrightarrow> \<exists>c. (b, c) \<in> R \<and> (a, c) \<in> R\<^sup>*"
+(* sledgehammer [isar_proof, isar_shrink_factor = 2] *)
+proof -
+ assume A1: "f c = Intg x"
+ assume A2: "\<forall>y. f b = Intg y \<longrightarrow> y \<noteq> x"
+ assume A3: "(a, b) \<in> R\<^sup>*"
+ assume A4: "(b, c) \<in> R\<^sup>*"
+ have "(R\<^sup>*) (a, b)" using A3 by (metis mem_def)
+ hence F1: "(a, b) \<in> R\<^sup>*" by (metis mem_def)
+ have "b \<noteq> c" using A1 A2 by metis
+ hence "\<exists>x\<^isub>1. (b, x\<^isub>1) \<in> R" using A4 by (metis converse_rtranclE)
+ thus "\<exists>c. (b, c) \<in> R \<and> (a, c) \<in> R\<^sup>*" using F1 by (metis transitive_closure_trans(6))
+qed
+
+lemma "\<lbrakk>f c = Intg x; \<forall>y. f b = Intg y \<longrightarrow> y \<noteq> x; (a, b) \<in> R\<^sup>*; (b, c) \<in> R\<^sup>*\<rbrakk>
+ \<Longrightarrow> \<exists>c. (b, c) \<in> R \<and> (a, c) \<in> R\<^sup>*"
+apply (erule_tac x = b in converse_rtranclE)
+ apply metis
+by (metis transitive_closure_trans(6))
+
+end