author | wenzelm |
Mon, 25 Feb 2013 12:17:50 +0100 | |
changeset 51272 | 9c8d63b4b6be |
parent 51143 | 0a2371e7ced3 |
child 55535 | 10194808430d |
permissions | -rw-r--r-- |
39157
b98909faaea8
more explicit HOL-Proofs sessions, including former ex/Hilbert_Classical.thy which works in parallel mode without the antiquotation option "margin" (which is still critical);
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parents:
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(* Title: HOL/Proofs/Extraction/Pigeonhole.thy |
17024 | 2 |
Author: Stefan Berghofer, TU Muenchen |
3 |
*) |
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header {* The pigeonhole principle *} |
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||
22737 | 7 |
theory Pigeonhole |
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two target language numeral types: integer and natural, as replacement for code_numeral;
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imports Util "~~/src/HOL/Library/Code_Target_Numeral" |
22737 | 9 |
begin |
17024 | 10 |
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text {* |
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We formalize two proofs of the pigeonhole principle, which lead |
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to extracted programs of quite different complexity. The original |
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formalization of these proofs in {\sc Nuprl} is due to |
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Aleksey Nogin \cite{Nogin-ENTCS-2000}. |
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This proof yields a polynomial program. |
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*} |
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theorem pigeonhole: |
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"\<And>f. (\<And>i. i \<le> Suc n \<Longrightarrow> f i \<le> n) \<Longrightarrow> \<exists>i j. i \<le> Suc n \<and> j < i \<and> f i = f j" |
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proof (induct n) |
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case 0 |
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hence "Suc 0 \<le> Suc 0 \<and> 0 < Suc 0 \<and> f (Suc 0) = f 0" by simp |
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17604 | 25 |
thus ?case by iprover |
17024 | 26 |
next |
27 |
case (Suc n) |
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{ |
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fix k |
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30 |
have |
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"k \<le> Suc (Suc n) \<Longrightarrow> |
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(\<And>i j. Suc k \<le> i \<Longrightarrow> i \<le> Suc (Suc n) \<Longrightarrow> j < i \<Longrightarrow> f i \<noteq> f j) \<Longrightarrow> |
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(\<exists>i j. i \<le> k \<and> j < i \<and> f i = f j)" |
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proof (induct k) |
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case 0 |
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let ?f = "\<lambda>i. if f i = Suc n then f (Suc (Suc n)) else f i" |
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37 |
have "\<not> (\<exists>i j. i \<le> Suc n \<and> j < i \<and> ?f i = ?f j)" |
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38 |
proof |
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parents:
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assume "\<exists>i j. i \<le> Suc n \<and> j < i \<and> ?f i = ?f j" |
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parents:
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then obtain i j where i: "i \<le> Suc n" and j: "j < i" |
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and f: "?f i = ?f j" by iprover |
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parents:
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from j have i_nz: "Suc 0 \<le> i" by simp |
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from i have iSSn: "i \<le> Suc (Suc n)" by simp |
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have S0SSn: "Suc 0 \<le> Suc (Suc n)" by simp |
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show False |
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proof cases |
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assume fi: "f i = Suc n" |
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show False |
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parents:
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proof cases |
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assume fj: "f j = Suc n" |
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from i_nz and iSSn and j have "f i \<noteq> f j" by (rule 0) |
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moreover from fi have "f i = f j" |
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parents:
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by (simp add: fj [symmetric]) |
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ultimately show ?thesis .. |
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next |
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parents:
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from i and j have "j < Suc (Suc n)" by simp |
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parents:
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with S0SSn and le_refl have "f (Suc (Suc n)) \<noteq> f j" |
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parents:
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by (rule 0) |
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parents:
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moreover assume "f j \<noteq> Suc n" |
69916a850301
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parents:
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diff
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|
60 |
with fi and f have "f (Suc (Suc n)) = f j" by simp |
69916a850301
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parents:
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ultimately show False .. |
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parents:
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qed |
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parents:
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63 |
next |
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parents:
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assume fi: "f i \<noteq> Suc n" |
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parents:
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|
65 |
show False |
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parents:
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|
66 |
proof cases |
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parents:
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from i have "i < Suc (Suc n)" by simp |
69916a850301
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parents:
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|
68 |
with S0SSn and le_refl have "f (Suc (Suc n)) \<noteq> f i" |
69916a850301
eliminated hard tabulators, guessing at each author's individual tab-width;
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parents:
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by (rule 0) |
69916a850301
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parents:
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moreover assume "f j = Suc n" |
69916a850301
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parents:
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|
71 |
with fi and f have "f (Suc (Suc n)) = f i" by simp |
69916a850301
eliminated hard tabulators, guessing at each author's individual tab-width;
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parents:
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diff
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72 |
ultimately show False .. |
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parents:
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|
73 |
next |
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parents:
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from i_nz and iSSn and j |
69916a850301
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parents:
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|
75 |
have "f i \<noteq> f j" by (rule 0) |
69916a850301
eliminated hard tabulators, guessing at each author's individual tab-width;
wenzelm
parents:
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diff
changeset
|
76 |
moreover assume "f j \<noteq> Suc n" |
69916a850301
eliminated hard tabulators, guessing at each author's individual tab-width;
wenzelm
parents:
29823
diff
changeset
|
77 |
with fi and f have "f i = f j" by simp |
69916a850301
eliminated hard tabulators, guessing at each author's individual tab-width;
wenzelm
parents:
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diff
changeset
|
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ultimately show False .. |
69916a850301
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parents:
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79 |
qed |
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parents:
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80 |
qed |
17024 | 81 |
qed |
82 |
moreover have "\<And>i. i \<le> Suc n \<Longrightarrow> ?f i \<le> n" |
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proof - |
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32960
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parents:
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84 |
fix i assume "i \<le> Suc n" |
69916a850301
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parents:
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|
85 |
hence i: "i < Suc (Suc n)" by simp |
69916a850301
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parents:
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|
86 |
have "f (Suc (Suc n)) \<noteq> f i" |
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parents:
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87 |
by (rule 0) (simp_all add: i) |
69916a850301
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parents:
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moreover have "f (Suc (Suc n)) \<le> Suc n" |
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parents:
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by (rule Suc) simp |
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parents:
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moreover from i have "i \<le> Suc (Suc n)" by simp |
69916a850301
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wenzelm
parents:
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91 |
hence "f i \<le> Suc n" by (rule Suc) |
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parents:
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92 |
ultimately show "?thesis i" |
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parents:
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by simp |
17024 | 94 |
qed |
95 |
hence "\<exists>i j. i \<le> Suc n \<and> j < i \<and> ?f i = ?f j" |
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32960
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parents:
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96 |
by (rule Suc) |
17024 | 97 |
ultimately show ?case .. |
98 |
next |
|
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case (Suc k) |
|
25418 | 100 |
from search [OF nat_eq_dec] show ?case |
17024 | 101 |
proof |
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parents:
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102 |
assume "\<exists>j<Suc k. f (Suc k) = f j" |
69916a850301
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parents:
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|
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thus ?case by (iprover intro: le_refl) |
17024 | 104 |
next |
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parents:
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105 |
assume nex: "\<not> (\<exists>j<Suc k. f (Suc k) = f j)" |
69916a850301
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parents:
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|
106 |
have "\<exists>i j. i \<le> k \<and> j < i \<and> f i = f j" |
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parents:
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107 |
proof (rule Suc) |
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from Suc show "k \<le> Suc (Suc n)" by simp |
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parents:
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|
109 |
fix i j assume k: "Suc k \<le> i" and i: "i \<le> Suc (Suc n)" |
69916a850301
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parents:
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and j: "j < i" |
69916a850301
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parents:
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111 |
show "f i \<noteq> f j" |
69916a850301
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parents:
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|
112 |
proof cases |
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113 |
assume eq: "i = Suc k" |
69916a850301
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parents:
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|
114 |
show ?thesis |
69916a850301
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|
115 |
proof |
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116 |
assume "f i = f j" |
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parents:
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117 |
hence "f (Suc k) = f j" by (simp add: eq) |
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parents:
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118 |
with nex and j and eq show False by iprover |
69916a850301
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119 |
qed |
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parents:
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|
120 |
next |
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121 |
assume "i \<noteq> Suc k" |
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122 |
with k have "Suc (Suc k) \<le> i" by simp |
69916a850301
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parents:
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123 |
thus ?thesis using i and j by (rule Suc) |
69916a850301
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124 |
qed |
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125 |
qed |
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|
126 |
thus ?thesis by (iprover intro: le_SucI) |
17024 | 127 |
qed |
128 |
qed |
|
129 |
} |
|
130 |
note r = this |
|
131 |
show ?case by (rule r) simp_all |
|
132 |
qed |
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133 |
||
134 |
text {* |
|
135 |
The following proof, although quite elegant from a mathematical point of view, |
|
136 |
leads to an exponential program: |
|
137 |
*} |
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138 |
||
139 |
theorem pigeonhole_slow: |
|
140 |
"\<And>f. (\<And>i. i \<le> Suc n \<Longrightarrow> f i \<le> n) \<Longrightarrow> \<exists>i j. i \<le> Suc n \<and> j < i \<and> f i = f j" |
|
141 |
proof (induct n) |
|
142 |
case 0 |
|
143 |
have "Suc 0 \<le> Suc 0" .. |
|
144 |
moreover have "0 < Suc 0" .. |
|
145 |
moreover from 0 have "f (Suc 0) = f 0" by simp |
|
17604 | 146 |
ultimately show ?case by iprover |
17024 | 147 |
next |
148 |
case (Suc n) |
|
25418 | 149 |
from search [OF nat_eq_dec] show ?case |
17024 | 150 |
proof |
151 |
assume "\<exists>j < Suc (Suc n). f (Suc (Suc n)) = f j" |
|
17604 | 152 |
thus ?case by (iprover intro: le_refl) |
17024 | 153 |
next |
154 |
assume "\<not> (\<exists>j < Suc (Suc n). f (Suc (Suc n)) = f j)" |
|
17604 | 155 |
hence nex: "\<forall>j < Suc (Suc n). f (Suc (Suc n)) \<noteq> f j" by iprover |
17024 | 156 |
let ?f = "\<lambda>i. if f i = Suc n then f (Suc (Suc n)) else f i" |
157 |
have "\<And>i. i \<le> Suc n \<Longrightarrow> ?f i \<le> n" |
|
158 |
proof - |
|
159 |
fix i assume i: "i \<le> Suc n" |
|
160 |
show "?thesis i" |
|
161 |
proof (cases "f i = Suc n") |
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162 |
case True |
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163 |
from i and nex have "f (Suc (Suc n)) \<noteq> f i" by simp |
69916a850301
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parents:
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164 |
with True have "f (Suc (Suc n)) \<noteq> Suc n" by simp |
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parents:
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165 |
moreover from Suc have "f (Suc (Suc n)) \<le> Suc n" by simp |
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parents:
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166 |
ultimately have "f (Suc (Suc n)) \<le> n" by simp |
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|
167 |
with True show ?thesis by simp |
17024 | 168 |
next |
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169 |
case False |
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170 |
from Suc and i have "f i \<le> Suc n" by simp |
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parents:
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171 |
with False show ?thesis by simp |
17024 | 172 |
qed |
173 |
qed |
|
174 |
hence "\<exists>i j. i \<le> Suc n \<and> j < i \<and> ?f i = ?f j" by (rule Suc) |
|
175 |
then obtain i j where i: "i \<le> Suc n" and ji: "j < i" and f: "?f i = ?f j" |
|
17604 | 176 |
by iprover |
17024 | 177 |
have "f i = f j" |
178 |
proof (cases "f i = Suc n") |
|
179 |
case True |
|
180 |
show ?thesis |
|
181 |
proof (cases "f j = Suc n") |
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32960
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182 |
assume "f j = Suc n" |
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183 |
with True show ?thesis by simp |
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next |
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assume "f j \<noteq> Suc n" |
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moreover from i ji nex have "f (Suc (Suc n)) \<noteq> f j" by simp |
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ultimately show ?thesis using True f by simp |
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qed |
189 |
next |
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case False |
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show ?thesis |
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proof (cases "f j = Suc n") |
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assume "f j = Suc n" |
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moreover from i nex have "f (Suc (Suc n)) \<noteq> f i" by simp |
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195 |
ultimately show ?thesis using False f by simp |
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assume "f j \<noteq> Suc n" |
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198 |
with False f show ?thesis by simp |
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qed |
200 |
qed |
|
201 |
moreover from i have "i \<le> Suc (Suc n)" by simp |
|
17604 | 202 |
ultimately show ?thesis using ji by iprover |
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qed |
204 |
qed |
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205 |
||
206 |
extract pigeonhole pigeonhole_slow |
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207 |
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208 |
text {* |
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The programs extracted from the above proofs look as follows: |
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@{thm [display] pigeonhole_def} |
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@{thm [display] pigeonhole_slow_def} |
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The program for searching for an element in an array is |
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@{thm [display,eta_contract=false] search_def} |
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The correctness statement for @{term "pigeonhole"} is |
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@{thm [display] pigeonhole_correctness [no_vars]} |
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216 |
||
217 |
In order to analyze the speed of the above programs, |
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218 |
we generate ML code from them. |
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219 |
*} |
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220 |
||
27982 | 221 |
instantiation nat :: default |
222 |
begin |
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223 |
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224 |
definition "default = (0::nat)" |
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225 |
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226 |
instance .. |
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227 |
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228 |
end |
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229 |
||
37678
0040bafffdef
"prod" and "sum" replace "*" and "+" respectively
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instantiation prod :: (default, default) default |
27982 | 231 |
begin |
232 |
||
233 |
definition "default = (default, default)" |
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234 |
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235 |
instance .. |
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236 |
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237 |
end |
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238 |
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definition |
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two target language numeral types: integer and natural, as replacement for code_numeral;
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"test n u = pigeonhole (nat_of_integer n) (\<lambda>m. m - 1)" |
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definition |
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242 |
"test' n u = pigeonhole_slow (nat_of_integer n) (\<lambda>m. m - 1)" |
22507 | 243 |
definition |
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244 |
"test'' u = pigeonhole 8 (List.nth [0, 1, 2, 3, 4, 5, 6, 3, 7, 8])" |
20837 | 245 |
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51272 | 246 |
ML_val "timeit (@{code test} 10)" |
247 |
ML_val "timeit (@{code test'} 10)" |
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248 |
ML_val "timeit (@{code test} 20)" |
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249 |
ML_val "timeit (@{code test'} 20)" |
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250 |
ML_val "timeit (@{code test} 25)" |
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251 |
ML_val "timeit (@{code test'} 25)" |
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252 |
ML_val "timeit (@{code test} 500)" |
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253 |
ML_val "timeit @{code test''}" |
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37287 | 254 |
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17024 | 255 |
end |
37287 | 256 |