src/HOL/Isar_examples/Puzzle.thy

theorems [cases type: bool] = case_split;

header {* An old chestnut *};

theory Puzzle = Main:;

text_raw {*
 \footnote{A question from ``Bundeswettbewerb Mathematik''.
 Original pen-and-paper proof due to Herbert Ehler; Isabelle tactic
 script by Tobias Nipkow.}
*};


subsection {* Generalized mathematical induction *};

text {*
 The following derived rule admits induction over some expression
 $f(x)$ wrt.\ the ${<}$ relation on natural numbers.
*};

lemma gen_less_induct:
  "(!!x. ALL y. f y < f x --> P y (f y) ==> P x (f x))
    ==> P x (f x :: nat)"
  (is "(!!x. ?H x ==> ?C x) ==> _");
proof -;
  assume asm: "!!x. ?H x ==> ?C x";
  {;
    fix k;
    have "ALL x. k = f x --> ?C x" (is "?Q k");
    proof (rule less_induct);
      fix k; assume hyp: "ALL m<k. ?Q m";
      show "?Q k";
      proof;
	fix x; show "k = f x --> ?C x";
	proof;
	  assume "k = f x";
	  with hyp; have "?H x"; by blast;
	  thus "?C x"; by (rule asm);
	qed;
      qed;
    qed;
  };
  thus "?C x"; by simp;
qed;


subsection {* The problem *};

text {*
 Given some function $f\colon \Nat \to \Nat$ such that $f \ap (f \ap
 n) < f \ap (\idt{Suc} \ap n)$ for all $n$.  Demonstrate that $f$ is
 the identity.
*};

consts f :: "nat => nat";
axioms f_ax: "f (f n) < f (Suc n)";

theorem "f n = n";
proof (rule order_antisym);
  txt {*
    Note that the generalized form of $n \le f \ap n$ is required
    later for monotonicity as well.
  *};
  show ge: "!!n. n <= f n";
  proof -;
    fix n;
    show "?thesis n" (is "?P n (f n)");
    proof (rule gen_less_induct [of f ?P]);
      fix n; assume hyp: "ALL m. f m < f n --> ?P m (f m)";
      show "?P n (f n)";
      proof (rule nat.exhaust);
	assume "n = 0"; thus ?thesis; by simp;
      next;
	fix m; assume n_Suc: "n = Suc m";
	from f_ax; have "f (f m) < f (Suc m)"; .;
	with hyp n_Suc; have "f m <= f (f m)"; by blast;
	also; from f_ax; have "... < f (Suc m)"; .;
	finally; have lt: "f m < f (Suc m)"; .;
	with hyp n_Suc; have "m <= f m"; by blast;
	also; note lt;
	finally; have "m < f (Suc m)"; .;
	thus "n <= f n"; by (simp only: n_Suc);
      qed;
    qed;
  qed;

  txt {*
    In order to show the other direction, we first establish
    monotonicity of $f$.
  *};
  have mono: "!!m n. m <= n --> f m <= f n";
  proof -;
    fix m n;
    show "?thesis m n" (is "?P n");
    proof (induct n);
      show "?P 0"; by simp;
      fix n; assume hyp: "?P n";
      show "?P (Suc n)";
      proof;
	assume "m <= Suc n";
	thus "f m <= f (Suc n)";
	proof (rule le_SucE);
	  assume "m <= n";
	  with hyp; have "f m <= f n"; ..;
	  also; from ge f_ax; have "... < f (Suc n)";
	    by (rule le_less_trans);
	  finally; show ?thesis; by simp;
	next;
	  assume "m = Suc n";
	  thus ?thesis; by simp;
	qed;
      qed;
    qed;
  qed;

  show "f n <= n";
  proof (rule leI);
    show "~ n < f n";
    proof;
      assume "n < f n";
      hence "Suc n <= f n"; by (rule Suc_leI);
      hence "f (Suc n) <= f (f n)"; by (rule mono [rulify]);
      also; have "... < f (Suc n)"; by (rule f_ax);
      finally; have "... < ..."; .; thus False; ..;
    qed;
  qed;
qed;

end;