--- a/ex/Puzzle.ML Tue Oct 24 14:59:17 1995 +0100
+++ /dev/null Thu Jan 01 00:00:00 1970 +0000
@@ -1,58 +0,0 @@
-(* Title: HOL/ex/puzzle.ML
- ID: $Id$
- Author: Tobias Nipkow
- Copyright 1993 TU Muenchen
-
-For puzzle.thy. A question from "Bundeswettbewerb Mathematik"
-
-Proof due to Herbert Ehler
-*)
-
-(*specialized form of induction needed below*)
-val prems = goal Nat.thy "[| P(0); !!n. P(Suc(n)) |] ==> !n.P(n)";
-by (EVERY1 [rtac (nat_induct RS allI), resolve_tac prems, resolve_tac prems]);
-qed "nat_exh";
-
-goal Puzzle.thy "! n. k=f(n) --> n <= f(n)";
-by (res_inst_tac [("n","k")] less_induct 1);
-by (rtac nat_exh 1);
-by (simp_tac nat_ss 1);
-by (rtac impI 1);
-by (rtac classical 1);
-by (dtac not_leE 1);
-by (subgoal_tac "f(na) <= f(f(na))" 1);
-by (best_tac (HOL_cs addIs [lessD,Puzzle.f_ax,le_less_trans,le_trans]) 1);
-by (fast_tac (HOL_cs addIs [Puzzle.f_ax]) 1);
-bind_thm("lemma", result() RS spec RS mp);
-
-goal Puzzle.thy "n <= f(n)";
-by (fast_tac (HOL_cs addIs [lemma]) 1);
-qed "lemma1";
-
-goal Puzzle.thy "f(n) < f(Suc(n))";
-by (fast_tac (HOL_cs addIs [Puzzle.f_ax,le_less_trans,lemma1]) 1);
-qed "lemma2";
-
-val prems = goal Puzzle.thy "(!!n.f(n) <= f(Suc(n))) ==> m<n --> f(m) <= f(n)";
-by (res_inst_tac[("n","n")]nat_induct 1);
-by (simp_tac nat_ss 1);
-by (simp_tac nat_ss 1);
-by (fast_tac (HOL_cs addIs (le_trans::prems)) 1);
-bind_thm("mono_lemma1", result() RS mp);
-
-val [p1,p2] = goal Puzzle.thy
- "[| !! n. f(n)<=f(Suc(n)); m<=n |] ==> f(m) <= f(n)";
-by (rtac (p2 RS le_imp_less_or_eq RS disjE) 1);
-by (etac (p1 RS mono_lemma1) 1);
-by (fast_tac (HOL_cs addIs [le_refl]) 1);
-qed "mono_lemma";
-
-val prems = goal Puzzle.thy "m <= n ==> f(m) <= f(n)";
-by (fast_tac (HOL_cs addIs ([mono_lemma,less_imp_le,lemma2]@prems)) 1);
-qed "f_mono";
-
-goal Puzzle.thy "f(n) = n";
-by (rtac le_anti_sym 1);
-by (rtac lemma1 2);
-by (fast_tac (HOL_cs addIs [Puzzle.f_ax,leI] addDs [leD,f_mono,lessD]) 1);
-result();