src/HOL/ex/InductiveInvariant_examples.thy
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(*  ID:         $Id$
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    Author:	Sava Krsti\'{c} and John Matthews
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*)
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header {* Example use if an inductive invariant to solve termination conditions *}
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theory InductiveInvariant_examples imports InductiveInvariant  begin
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text "A simple example showing how to use an inductive invariant
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      to solve termination conditions generated by recdef on
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      nested recursive function definitions."
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consts g :: "nat => nat"
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recdef (permissive) g "less_than"
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  "g 0 = 0"
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  "g (Suc n) = g (g n)"
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text "We can prove the unsolved termination condition for
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      g by showing it is an inductive invariant."
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recdef_tc g_tc[simp]: g
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apply (rule allI)
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apply (rule_tac x=n in tfl_indinv_wfrec [OF g_def])
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apply (auto simp add: indinv_def split: nat.split)
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apply (frule_tac x=nat in spec)
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apply (drule_tac x="f nat" in spec)
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by auto
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text "This declaration invokes Isabelle's simplifier to
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      remove any termination conditions before adding
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      g's rules to the simpset."
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declare g.simps [simplified, simp]
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text "This is an example where the termination condition generated
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      by recdef is not itself an inductive invariant."
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consts g' :: "nat => nat"
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recdef (permissive) g' "less_than"
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  "g' 0 = 0"
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  "g' (Suc n) = g' n + g' (g' n)"
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thm g'.simps
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text "The strengthened inductive invariant is as follows
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      (this invariant also works for the first example above):"
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lemma g'_inv: "g' n = 0"
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thm tfl_indinv_wfrec [OF g'_def]
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apply (rule_tac x=n in tfl_indinv_wfrec [OF g'_def])
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by (auto simp add: indinv_def split: nat.split)
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recdef_tc g'_tc[simp]: g'
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by (simp add: g'_inv)
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text "Now we can remove the termination condition from
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      the rules for g' ."
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thm g'.simps [simplified]
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text {* Sometimes a recursive definition is partial, that is, it
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        is only meant to be invoked on "good" inputs. As a contrived
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        example, we will define a new version of g that is only
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        well defined for even inputs greater than zero. *}
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consts g_even :: "nat => nat"
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recdef (permissive) g_even "less_than"
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  "g_even (Suc (Suc 0)) = 3"
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  "g_even n = g_even (g_even (n - 2) - 1)"
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text "We can prove a conditional version of the unsolved termination
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      condition for @{term g_even} by proving a stronger inductive invariant."
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lemma g_even_indinv: "\<exists>k. n = Suc (Suc (2*k)) ==> g_even n = 3"
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apply (rule_tac D="{n. \<exists>k. n = Suc (Suc (2*k))}" and x=n in tfl_indinv_on_wfrec [OF g_even_def])
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apply (auto simp add: indinv_on_def split: nat.split)
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by (case_tac ka, auto)
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text "Now we can prove that the second recursion equation for @{term g_even}
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      holds, provided that n is an even number greater than two."
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theorem g_even_n: "\<exists>k. n = 2*k + 4 ==> g_even n = g_even (g_even (n - 2) - 1)"
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apply (subgoal_tac "(\<exists>k. n - 2 = 2*k + 2) & (\<exists>k. n = 2*k + 2)")
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by (auto simp add: g_even_indinv, arith)
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text "McCarthy's ninety-one function. This function requires a
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      non-standard measure to prove termination."
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consts ninety_one :: "nat => nat"
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recdef (permissive) ninety_one "measure (%n. 101 - n)"
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  "ninety_one x = (if 100 < x
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                     then x - 10
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                     else (ninety_one (ninety_one (x+11))))"
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text "To discharge the termination condition, we will prove
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      a strengthened inductive invariant:
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         S x y == x < y + 11"
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lemma ninety_one_inv: "n < ninety_one n + 11"
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apply (rule_tac x=n in tfl_indinv_wfrec [OF ninety_one_def])
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apply force
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apply (auto simp add: indinv_def)
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apply (frule_tac x="x+11" in spec)
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apply (frule_tac x="f (x + 11)" in spec)
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by arith
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text "Proving the termination condition using the
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      strengthened inductive invariant."
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recdef_tc ninety_one_tc[rule_format]: ninety_one
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apply clarify
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by (cut_tac n="x+11" in ninety_one_inv, arith)
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text "Now we can remove the termination condition from
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      the simplification rule for @{term ninety_one}."
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theorem def_ninety_one:
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"ninety_one x = (if 100 < x
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                   then x - 10
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                   else ninety_one (ninety_one (x+11)))"
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by (subst ninety_one.simps,
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    simp add: ninety_one_tc)
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end