src/HOL/ex/Predicate_Compile_ex.thy
author haftmann
Tue, 14 Jul 2009 16:27:32 +0200
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permissions -rw-r--r--
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theory Predicate_Compile_ex
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imports Complex_Main Predicate_Compile
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begin
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inductive even :: "nat \<Rightarrow> bool" and odd :: "nat \<Rightarrow> bool" where
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    "even 0"
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  | "even n \<Longrightarrow> odd (Suc n)"
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  | "odd n \<Longrightarrow> even (Suc n)"
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code_pred even .
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thm odd.equation
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thm even.equation
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values "{x. even 2}"
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values "{x. odd 2}"
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values 10 "{n. even n}"
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values 10 "{n. odd n}"
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inductive append :: "'a list \<Rightarrow> 'a list \<Rightarrow> 'a list \<Rightarrow> bool" where
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    append_Nil: "append [] xs xs"
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  | append_Cons: "append xs ys zs \<Longrightarrow> append (x # xs) ys (x # zs)"
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inductive rev
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where
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"rev [] []"
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| "rev xs xs' ==> append xs' [x] ys ==> rev (x#xs) ys"
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code_pred rev .
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thm append.equation
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values "{(ys, xs). append xs ys [0, Suc 0, 2]}"
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values "{zs. append [0, Suc 0, 2] [17, 8] zs}"
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values "{ys. append [0, Suc 0, 2] ys [0, Suc 0, 2, 17, 0,5]}"
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inductive partition :: "('a \<Rightarrow> bool) \<Rightarrow> 'a list \<Rightarrow> 'a list \<Rightarrow> 'a list \<Rightarrow> bool"
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  for f where
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    "partition f [] [] []"
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  | "f x \<Longrightarrow> partition f xs ys zs \<Longrightarrow> partition f (x # xs) (x # ys) zs"
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  | "\<not> f x \<Longrightarrow> partition f xs ys zs \<Longrightarrow> partition f (x # xs) ys (x # zs)"
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(* FIXME: correct handling of parameters *)
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(*
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ML {* reset Predicate_Compile.do_proofs *}
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code_pred partition .
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thm partition.equation
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ML {* set Predicate_Compile.do_proofs *}
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*)
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(* TODO: requires to handle abstractions in parameter positions correctly *)
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(*FIXME values 10 "{(ys, zs). partition (\<lambda>n. n mod 2 = 0)
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  [0, Suc 0, 2, 3, 4, 5, 6, 7] ys zs}" *)
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lemma [code_pred_intros]:
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"r a b ==> tranclp r a b"
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"r a b ==> tranclp r b c ==> tranclp r a c"
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by auto
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(* Setup requires quick and dirty proof *)
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(*
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code_pred tranclp
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proof -
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  case tranclp
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  from this converse_tranclpE[OF this(1)] show thesis by metis
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qed
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thm tranclp.equation
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*)
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inductive succ :: "nat \<Rightarrow> nat \<Rightarrow> bool" where
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    "succ 0 1"
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  | "succ m n \<Longrightarrow> succ (Suc m) (Suc n)"
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code_pred succ .
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thm succ.equation
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(* FIXME: why does this not terminate? *)
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(*
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values 20 "{n. tranclp succ 10 n}"
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values "{n. tranclp succ n 10}"
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values 20 "{(n, m). tranclp succ n m}"
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*)
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end