author | wenzelm |
Sun, 30 Nov 2008 14:43:29 +0100 | |
changeset 28917 | 20f43e0e0958 |
parent 25314 | 5eaf3e8b50a4 |
permissions | -rw-r--r-- |
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1 |
(* Title: HOL/Library/SCT_Misc.thy |
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ID: $Id$ |
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Author: Alexander Krauss, TU Muenchen |
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*) |
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header {* Miscellaneous Tools for Size-Change Termination *} |
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7 |
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theory Misc_Tools |
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imports Main |
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begin |
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|
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subsection {* Searching in lists *} |
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|
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fun index_of :: "'a list \<Rightarrow> 'a \<Rightarrow> nat" |
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where |
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"index_of [] c = 0" |
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| "index_of (x#xs) c = (if x = c then 0 else Suc (index_of xs c))" |
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lemma index_of_member: |
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"(x \<in> set l) \<Longrightarrow> (l ! index_of l x = x)" |
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by (induct l) auto |
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lemma index_of_length: |
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"(x \<in> set l) = (index_of l x < length l)" |
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25 |
by (induct l) auto |
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subsection {* Some reasoning tools *} |
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lemma three_cases: |
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30 |
assumes "a1 \<Longrightarrow> thesis" |
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31 |
assumes "a2 \<Longrightarrow> thesis" |
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32 |
assumes "a3 \<Longrightarrow> thesis" |
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assumes "\<And>R. \<lbrakk>a1 \<Longrightarrow> R; a2 \<Longrightarrow> R; a3 \<Longrightarrow> R\<rbrakk> \<Longrightarrow> R" |
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34 |
shows "thesis" |
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using assms |
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by auto |
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subsection {* Sequences *} |
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types |
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'a sequence = "nat \<Rightarrow> 'a" |
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subsubsection {* Increasing sequences *} |
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definition |
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increasing :: "(nat \<Rightarrow> nat) \<Rightarrow> bool" where |
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"increasing s = (\<forall>i j. i < j \<longrightarrow> s i < s j)" |
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lemma increasing_strict: |
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52 |
assumes "increasing s" |
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assumes "i < j" |
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shows "s i < s j" |
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using assms |
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unfolding increasing_def by simp |
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lemma increasing_weak: |
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59 |
assumes "increasing s" |
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60 |
assumes "i \<le> j" |
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61 |
shows "s i \<le> s j" |
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62 |
using assms increasing_strict[of s i j] |
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by (cases "i < j") auto |
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lemma increasing_inc: |
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66 |
assumes "increasing s" |
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67 |
shows "n \<le> s n" |
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proof (induct n) |
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case 0 then show ?case by simp |
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70 |
next |
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case (Suc n) |
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72 |
with increasing_strict [OF `increasing s`, of n "Suc n"] |
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73 |
show ?case by auto |
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qed |
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75 |
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lemma increasing_bij: |
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77 |
assumes [simp]: "increasing s" |
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78 |
shows "(s i < s j) = (i < j)" |
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79 |
proof |
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80 |
assume "s i < s j" |
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81 |
show "i < j" |
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82 |
proof (rule classical) |
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83 |
assume "\<not> ?thesis" |
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hence "j \<le> i" by arith |
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85 |
with increasing_weak have "s j \<le> s i" by simp |
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86 |
with `s i < s j` show ?thesis by simp |
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87 |
qed |
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88 |
qed (simp add:increasing_strict) |
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89 |
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90 |
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91 |
subsubsection {* Sections induced by an increasing sequence *} |
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92 |
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abbreviation |
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94 |
"section s i == {s i ..< s (Suc i)}" |
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95 |
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definition |
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97 |
"section_of s n = (LEAST i. n < s (Suc i))" |
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98 |
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99 |
lemma section_help: |
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100 |
assumes "increasing s" |
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101 |
shows "\<exists>i. n < s (Suc i)" |
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102 |
proof - |
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103 |
have "n \<le> s n" |
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104 |
using `increasing s` by (rule increasing_inc) |
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105 |
also have "\<dots> < s (Suc n)" |
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106 |
using `increasing s` increasing_strict by simp |
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107 |
finally show ?thesis .. |
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108 |
qed |
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109 |
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110 |
lemma section_of2: |
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111 |
assumes "increasing s" |
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112 |
shows "n < s (Suc (section_of s n))" |
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113 |
unfolding section_of_def |
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114 |
by (rule LeastI_ex) (rule section_help [OF `increasing s`]) |
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115 |
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116 |
lemma section_of1: |
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117 |
assumes [simp, intro]: "increasing s" |
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118 |
assumes "s i \<le> n" |
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119 |
shows "s (section_of s n) \<le> n" |
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120 |
proof (rule classical) |
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121 |
let ?m = "section_of s n" |
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122 |
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123 |
assume "\<not> ?thesis" |
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124 |
hence a: "n < s ?m" by simp |
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125 |
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126 |
have nonzero: "?m \<noteq> 0" |
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127 |
proof |
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128 |
assume "?m = 0" |
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129 |
from increasing_weak have "s 0 \<le> s i" by simp |
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130 |
also note `\<dots> \<le> n` |
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131 |
finally show False using `?m = 0` `n < s ?m` by simp |
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132 |
qed |
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133 |
with a have "n < s (Suc (?m - 1))" by simp |
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with Least_le have "?m \<le> ?m - 1" |
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135 |
unfolding section_of_def . |
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136 |
with nonzero show ?thesis by simp |
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137 |
qed |
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138 |
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139 |
lemma section_of_known: |
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140 |
assumes [simp]: "increasing s" |
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141 |
assumes in_sect: "k \<in> section s i" |
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142 |
shows "section_of s k = i" (is "?s = i") |
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143 |
proof (rule classical) |
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144 |
assume "\<not> ?thesis" |
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145 |
|
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146 |
hence "?s < i \<or> ?s > i" by arith |
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147 |
thus ?thesis |
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148 |
proof |
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149 |
assume "?s < i" |
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150 |
hence "Suc ?s \<le> i" by simp |
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151 |
with increasing_weak have "s (Suc ?s) \<le> s i" by simp |
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152 |
moreover have "k < s (Suc ?s)" using section_of2 by simp |
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153 |
moreover from in_sect have "s i \<le> k" by simp |
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154 |
ultimately show ?thesis by simp |
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155 |
next |
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156 |
assume "i < ?s" hence "Suc i \<le> ?s" by simp |
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157 |
with increasing_weak have "s (Suc i) \<le> s ?s" by simp |
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|
158 |
moreover |
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159 |
from in_sect have "s i \<le> k" by simp |
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160 |
with section_of1 have "s ?s \<le> k" by simp |
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161 |
moreover from in_sect have "k < s (Suc i)" by simp |
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162 |
ultimately show ?thesis by simp |
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163 |
qed |
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164 |
qed |
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|
165 |
|
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166 |
lemma in_section_of: |
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167 |
assumes "increasing s" |
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168 |
assumes "s i \<le> k" |
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169 |
shows "k \<in> section s (section_of s k)" |
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170 |
using assms |
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171 |
by (auto intro:section_of1 section_of2) |
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172 |
|
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173 |
end |