diff -r 000000000000 -r 7949f97df77a ex/Puzzle.ML --- /dev/null Thu Jan 01 00:00:00 1970 +0000 +++ b/ex/Puzzle.ML Thu Sep 16 12:21:07 1993 +0200 @@ -0,0 +1,58 @@ +(* 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]); +val nat_exh = result(); + +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); +val lemma = result() RS spec RS mp; + +goal Puzzle.thy "n <= f(n)"; +by (fast_tac (HOL_cs addIs [lemma]) 1); +val lemma1 = result(); + +goal Puzzle.thy "f(n) < f(Suc(n))"; +by (fast_tac (HOL_cs addIs [Puzzle.f_ax,le_less_trans,lemma1]) 1); +val lemma2 = result(); + +val prems = goal Puzzle.thy "(!!n.f(n) <= f(Suc(n))) ==> m 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); +val 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); +val mono_lemma = result(); + +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); +val f_mono = result(); + +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();