|
1 |
|
2 header {* An old chestnut *}; |
|
3 |
|
4 theory Puzzle = Main:; |
|
5 |
|
6 text_raw {* |
|
7 \footnote{A question from ``Bundeswettbewerb Mathematik''. |
|
8 Original pen-and-paper proof due to Herbert Ehler; Isabelle tactic |
|
9 script by Tobias Nipkow.} |
|
10 *}; |
|
11 |
|
12 |
|
13 subsection {* Generalized mathematical induction *}; |
|
14 |
|
15 text {* |
|
16 The following derived rule admits induction over some expression |
|
17 $f(x)$ wrt.\ the ${<}$ relation on natural numbers. |
|
18 *}; |
|
19 |
|
20 lemma gen_less_induct: |
|
21 "(!!x. ALL y. f y < f x --> P y (f y) ==> P x (f x)) |
|
22 ==> P x (f x :: nat)" |
|
23 (is "(!!x. ?H x ==> ?C x) ==> _"); |
|
24 proof -; |
|
25 assume asm: "!!x. ?H x ==> ?C x"; |
|
26 {{; |
|
27 fix k; |
|
28 have "ALL x. k = f x --> ?C x" (is "?Q k"); |
|
29 proof (rule less_induct); |
|
30 fix k; assume hyp: "ALL m<k. ?Q m"; |
|
31 show "?Q k"; |
|
32 proof; |
|
33 fix x; show "k = f x --> ?C x"; |
|
34 proof; |
|
35 assume "k = f x"; |
|
36 with hyp; have "?H x"; by blast; |
|
37 thus "?C x"; by (rule asm); |
|
38 qed; |
|
39 qed; |
|
40 qed; |
|
41 }}; |
|
42 thus "?C x"; by simp; |
|
43 qed; |
|
44 |
|
45 |
|
46 subsection {* The problem *}; |
|
47 |
|
48 text {* |
|
49 Given some function $f\colon \Nat \to \Nat$ such that $f \ap (f \ap |
|
50 n) < f \ap (\idt{Suc} \ap n)$ for all $n$. Demonstrate that $f$ is |
|
51 the identity. |
|
52 *}; |
|
53 |
|
54 consts f :: "nat => nat"; |
|
55 axioms f_ax: "f (f n) < f (Suc n)"; |
|
56 |
|
57 theorem "f n = n"; |
|
58 proof (rule order_antisym); |
|
59 txt {* |
|
60 Note that the generalized form of $n \le f \ap n$ is required |
|
61 later for monotonicity as well. |
|
62 *}; |
|
63 show ge: "!!n. n <= f n"; |
|
64 proof -; |
|
65 fix n; |
|
66 show "?thesis n" (is "?P n (f n)"); |
|
67 proof (rule gen_less_induct [of f ?P]); |
|
68 fix n; assume hyp: "ALL m. f m < f n --> ?P m (f m)"; |
|
69 show "?P n (f n)"; |
|
70 proof (rule nat.exhaust); |
|
71 assume "n = 0"; thus ?thesis; by simp; |
|
72 next; |
|
73 fix m; assume n_Suc: "n = Suc m"; |
|
74 from f_ax; have "f (f m) < f (Suc m)"; .; |
|
75 with hyp n_Suc; have "f m <= f (f m)"; by blast; |
|
76 also; from f_ax; have "... < f (Suc m)"; .; |
|
77 finally; have lt: "f m < f (Suc m)"; .; |
|
78 with hyp n_Suc; have "m <= f m"; by blast; |
|
79 also; note lt; |
|
80 finally; have "m < f (Suc m)"; .; |
|
81 thus "n <= f n"; by (simp only: n_Suc); |
|
82 qed; |
|
83 qed; |
|
84 qed; |
|
85 |
|
86 txt {* |
|
87 In order to show the other direction, we first establish |
|
88 monotonicity of $f$. |
|
89 *}; |
|
90 have mono: "!!m n. m <= n --> f m <= f n"; |
|
91 proof -; |
|
92 fix m n; |
|
93 show "?thesis m n" (is "?P n"); |
|
94 proof (induct n); |
|
95 show "?P 0"; by simp; |
|
96 fix n; assume hyp: "?P n"; |
|
97 show "?P (Suc n)"; |
|
98 proof; |
|
99 assume "m <= Suc n"; |
|
100 thus "f m <= f (Suc n)"; |
|
101 proof (rule le_SucE); |
|
102 assume "m <= n"; |
|
103 with hyp; have "f m <= f n"; ..; |
|
104 also; from ge f_ax; have "... < f (Suc n)"; |
|
105 by (rule le_less_trans); |
|
106 finally; show ?thesis; by simp; |
|
107 next; |
|
108 assume "m = Suc n"; |
|
109 thus ?thesis; by simp; |
|
110 qed; |
|
111 qed; |
|
112 qed; |
|
113 qed; |
|
114 |
|
115 show "f n <= n"; |
|
116 proof (rule leI); |
|
117 show "~ n < f n"; |
|
118 proof; |
|
119 assume "n < f n"; |
|
120 hence "Suc n <= f n"; by (rule Suc_leI); |
|
121 hence "f (Suc n) <= f (f n)"; by (rule mono [rulify]); |
|
122 also; have "... < f (Suc n)"; by (rule f_ax); |
|
123 finally; have "... < ..."; .; thus False; ..; |
|
124 qed; |
|
125 qed; |
|
126 qed; |
|
127 |
|
128 end; |