author  wenzelm 
Sat, 07 Apr 2012 16:41:59 +0200  
changeset 47389  e8552cba702d 
parent 47317  432b29a96f61 
child 47432  e1576d13e933 
permissions  rwrr 
23465  1 
(* Title: HOL/Presburger.thy 
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Author: Amine Chaieb, TU Muenchen 

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*) 

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23472  5 
header {* Decision Procedure for Presburger Arithmetic *} 
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23465  7 
theory Presburger 
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imports Groebner_Basis Set_Interval 
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uses 
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"Tools/Qelim/qelim.ML" 
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"Tools/Qelim/cooper_procedure.ML" 
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("Tools/Qelim/cooper.ML") 
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begin 

14 

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subsection{* The @{text "\<infinity>"} and @{text "+\<infinity>"} Properties *} 

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lemma minf: 

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"\<lbrakk>\<exists>(z ::'a::linorder).\<forall>x<z. P x = P' x; \<exists>z.\<forall>x<z. Q x = Q' x\<rbrakk> 

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\<Longrightarrow> \<exists>z.\<forall>x<z. (P x \<and> Q x) = (P' x \<and> Q' x)" 

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"\<lbrakk>\<exists>(z ::'a::linorder).\<forall>x<z. P x = P' x; \<exists>z.\<forall>x<z. Q x = Q' x\<rbrakk> 

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\<Longrightarrow> \<exists>z.\<forall>x<z. (P x \<or> Q x) = (P' x \<or> Q' x)" 

22 
"\<exists>(z ::'a::{linorder}).\<forall>x<z.(x = t) = False" 

23 
"\<exists>(z ::'a::{linorder}).\<forall>x<z.(x \<noteq> t) = True" 

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"\<exists>(z ::'a::{linorder}).\<forall>x<z.(x < t) = True" 

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"\<exists>(z ::'a::{linorder}).\<forall>x<z.(x \<le> t) = True" 

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"\<exists>(z ::'a::{linorder}).\<forall>x<z.(x > t) = False" 

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"\<exists>(z ::'a::{linorder}).\<forall>x<z.(x \<ge> t) = False" 

45425  28 
"\<exists>z.\<forall>(x::'b::{linorder,plus,Rings.dvd})<z. (d dvd x + s) = (d dvd x + s)" 
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"\<exists>z.\<forall>(x::'b::{linorder,plus,Rings.dvd})<z. (\<not> d dvd x + s) = (\<not> d dvd x + s)" 

23465  30 
"\<exists>z.\<forall>x<z. F = F" 
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by ((erule exE, erule exE,rule_tac x="min z za" in exI,simp)+, (rule_tac x="t" in exI,fastforce)+) simp_all 
23465  32 

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lemma pinf: 

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"\<lbrakk>\<exists>(z ::'a::linorder).\<forall>x>z. P x = P' x; \<exists>z.\<forall>x>z. Q x = Q' x\<rbrakk> 

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\<Longrightarrow> \<exists>z.\<forall>x>z. (P x \<and> Q x) = (P' x \<and> Q' x)" 

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"\<lbrakk>\<exists>(z ::'a::linorder).\<forall>x>z. P x = P' x; \<exists>z.\<forall>x>z. Q x = Q' x\<rbrakk> 

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\<Longrightarrow> \<exists>z.\<forall>x>z. (P x \<or> Q x) = (P' x \<or> Q' x)" 

38 
"\<exists>(z ::'a::{linorder}).\<forall>x>z.(x = t) = False" 

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"\<exists>(z ::'a::{linorder}).\<forall>x>z.(x \<noteq> t) = True" 

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"\<exists>(z ::'a::{linorder}).\<forall>x>z.(x < t) = False" 

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"\<exists>(z ::'a::{linorder}).\<forall>x>z.(x \<le> t) = False" 

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"\<exists>(z ::'a::{linorder}).\<forall>x>z.(x > t) = True" 

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"\<exists>(z ::'a::{linorder}).\<forall>x>z.(x \<ge> t) = True" 

45425  44 
"\<exists>z.\<forall>(x::'b::{linorder,plus,Rings.dvd})>z. (d dvd x + s) = (d dvd x + s)" 
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"\<exists>z.\<forall>(x::'b::{linorder,plus,Rings.dvd})>z. (\<not> d dvd x + s) = (\<not> d dvd x + s)" 

23465  46 
"\<exists>z.\<forall>x>z. F = F" 
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by ((erule exE, erule exE,rule_tac x="max z za" in exI,simp)+,(rule_tac x="t" in exI,fastforce)+) simp_all 
23465  48 

49 
lemma inf_period: 

50 
"\<lbrakk>\<forall>x k. P x = P (x  k*D); \<forall>x k. Q x = Q (x  k*D)\<rbrakk> 

51 
\<Longrightarrow> \<forall>x k. (P x \<and> Q x) = (P (x  k*D) \<and> Q (x  k*D))" 

52 
"\<lbrakk>\<forall>x k. P x = P (x  k*D); \<forall>x k. Q x = Q (x  k*D)\<rbrakk> 

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\<Longrightarrow> \<forall>x k. (P x \<or> Q x) = (P (x  k*D) \<or> Q (x  k*D))" 

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"(d::'a::{comm_ring,Rings.dvd}) dvd D \<Longrightarrow> \<forall>x k. (d dvd x + t) = (d dvd (x  k*D) + t)" 
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"(d::'a::{comm_ring,Rings.dvd}) dvd D \<Longrightarrow> \<forall>x k. (\<not>d dvd x + t) = (\<not>d dvd (x  k*D) + t)" 
23465  56 
"\<forall>x k. F = F" 
29667  57 
apply (auto elim!: dvdE simp add: algebra_simps) 
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unfolding mult_assoc [symmetric] left_distrib [symmetric] left_diff_distrib [symmetric] 
27668  59 
unfolding dvd_def mult_commute [of d] 
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by auto 

23465  61 

23472  62 
subsection{* The A and B sets *} 
23465  63 
lemma bset: 
64 
"\<lbrakk>\<forall>x.(\<forall>j \<in> {1 .. D}. \<forall>b\<in>B. x \<noteq> b + j)\<longrightarrow> P x \<longrightarrow> P(x  D) ; 

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\<forall>x.(\<forall>j\<in>{1 .. D}. \<forall>b\<in>B. x \<noteq> b + j)\<longrightarrow> Q x \<longrightarrow> Q(x  D)\<rbrakk> \<Longrightarrow> 

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\<forall>x.(\<forall>j\<in>{1 .. D}. \<forall>b\<in>B. x \<noteq> b + j) \<longrightarrow> (P x \<and> Q x) \<longrightarrow> (P(x  D) \<and> Q (x  D))" 

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"\<lbrakk>\<forall>x.(\<forall>j\<in>{1 .. D}. \<forall>b\<in>B. x \<noteq> b + j)\<longrightarrow> P x \<longrightarrow> P(x  D) ; 

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\<forall>x.(\<forall>j\<in>{1 .. D}. \<forall>b\<in>B. x \<noteq> b + j)\<longrightarrow> Q x \<longrightarrow> Q(x  D)\<rbrakk> \<Longrightarrow> 

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\<forall>x.(\<forall>j\<in>{1 .. D}. \<forall>b\<in>B. x \<noteq> b + j)\<longrightarrow> (P x \<or> Q x) \<longrightarrow> (P(x  D) \<or> Q (x  D))" 

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"\<lbrakk>D>0; t  1\<in> B\<rbrakk> \<Longrightarrow> (\<forall>x.(\<forall>j\<in>{1 .. D}. \<forall>b\<in>B. x \<noteq> b + j)\<longrightarrow> (x = t) \<longrightarrow> (x  D = t))" 

71 
"\<lbrakk>D>0 ; t \<in> B\<rbrakk> \<Longrightarrow>(\<forall>(x::int).(\<forall>j\<in>{1 .. D}. \<forall>b\<in>B. x \<noteq> b + j)\<longrightarrow> (x \<noteq> t) \<longrightarrow> (x  D \<noteq> t))" 

72 
"D>0 \<Longrightarrow> (\<forall>(x::int).(\<forall>j\<in>{1 .. D}. \<forall>b\<in>B. x \<noteq> b + j)\<longrightarrow> (x < t) \<longrightarrow> (x  D < t))" 

73 
"D>0 \<Longrightarrow> (\<forall>(x::int).(\<forall>j\<in>{1 .. D}. \<forall>b\<in>B. x \<noteq> b + j)\<longrightarrow> (x \<le> t) \<longrightarrow> (x  D \<le> t))" 

74 
"\<lbrakk>D>0 ; t \<in> B\<rbrakk> \<Longrightarrow>(\<forall>(x::int).(\<forall>j\<in>{1 .. D}. \<forall>b\<in>B. x \<noteq> b + j)\<longrightarrow> (x > t) \<longrightarrow> (x  D > t))" 

75 
"\<lbrakk>D>0 ; t  1 \<in> B\<rbrakk> \<Longrightarrow>(\<forall>(x::int).(\<forall>j\<in>{1 .. D}. \<forall>b\<in>B. x \<noteq> b + j)\<longrightarrow> (x \<ge> t) \<longrightarrow> (x  D \<ge> t))" 

76 
"d dvd D \<Longrightarrow>(\<forall>(x::int).(\<forall>j\<in>{1 .. D}. \<forall>b\<in>B. x \<noteq> b + j)\<longrightarrow> (d dvd x+t) \<longrightarrow> (d dvd (x  D) + t))" 

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"d dvd D \<Longrightarrow>(\<forall>(x::int).(\<forall>j\<in>{1 .. D}. \<forall>b\<in>B. x \<noteq> b + j)\<longrightarrow> (\<not>d dvd x+t) \<longrightarrow> (\<not> d dvd (x  D) + t))" 

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"\<forall>x.(\<forall>j\<in>{1 .. D}. \<forall>b\<in>B. x \<noteq> b + j) \<longrightarrow> F \<longrightarrow> F" 

79 
proof (blast, blast) 

80 
assume dp: "D > 0" and tB: "t  1\<in> B" 

81 
show "(\<forall>x.(\<forall>j\<in>{1 .. D}. \<forall>b\<in>B. x \<noteq> b + j)\<longrightarrow> (x = t) \<longrightarrow> (x  D = t))" 

27668  82 
apply (rule allI, rule impI,erule ballE[where x="1"],erule ballE[where x="t  1"]) 
83 
apply algebra using dp tB by simp_all 

23465  84 
next 
85 
assume dp: "D > 0" and tB: "t \<in> B" 

86 
show "(\<forall>x.(\<forall>j\<in>{1 .. D}. \<forall>b\<in>B. x \<noteq> b + j)\<longrightarrow> (x \<noteq> t) \<longrightarrow> (x  D \<noteq> t))" 

87 
apply (rule allI, rule impI,erule ballE[where x="D"],erule ballE[where x="t"]) 

27668  88 
apply algebra 
23465  89 
using dp tB by simp_all 
90 
next 

91 
assume dp: "D > 0" thus "(\<forall>x.(\<forall>j\<in>{1 .. D}. \<forall>b\<in>B. x \<noteq> b + j)\<longrightarrow> (x < t) \<longrightarrow> (x  D < t))" by arith 

92 
next 

93 
assume dp: "D > 0" thus "\<forall>x.(\<forall>j\<in>{1 .. D}. \<forall>b\<in>B. x \<noteq> b + j)\<longrightarrow> (x \<le> t) \<longrightarrow> (x  D \<le> t)" by arith 

94 
next 

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assume dp: "D > 0" and tB:"t \<in> B" 

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{fix x assume nob: "\<forall>j\<in>{1 .. D}. \<forall>b\<in>B. x \<noteq> b + j" and g: "x > t" and ng: "\<not> (x  D) > t" 

97 
hence "x t \<le> D" and "1 \<le> x  t" by simp+ 

98 
hence "\<exists>j \<in> {1 .. D}. x  t = j" by auto 

29667  99 
hence "\<exists>j \<in> {1 .. D}. x = t + j" by (simp add: algebra_simps) 
23465  100 
with nob tB have "False" by simp} 
101 
thus "\<forall>x.(\<forall>j\<in>{1 .. D}. \<forall>b\<in>B. x \<noteq> b + j)\<longrightarrow> (x > t) \<longrightarrow> (x  D > t)" by blast 

102 
next 

103 
assume dp: "D > 0" and tB:"t  1\<in> B" 

104 
{fix x assume nob: "\<forall>j\<in>{1 .. D}. \<forall>b\<in>B. x \<noteq> b + j" and g: "x \<ge> t" and ng: "\<not> (x  D) \<ge> t" 

105 
hence "x  (t  1) \<le> D" and "1 \<le> x  (t  1)" by simp+ 

106 
hence "\<exists>j \<in> {1 .. D}. x  (t  1) = j" by auto 

29667  107 
hence "\<exists>j \<in> {1 .. D}. x = (t  1) + j" by (simp add: algebra_simps) 
23465  108 
with nob tB have "False" by simp} 
109 
thus "\<forall>x.(\<forall>j\<in>{1 .. D}. \<forall>b\<in>B. x \<noteq> b + j)\<longrightarrow> (x \<ge> t) \<longrightarrow> (x  D \<ge> t)" by blast 

110 
next 

111 
assume d: "d dvd D" 

27668  112 
{fix x assume H: "d dvd x + t" with d have "d dvd (x  D) + t" by algebra} 
23465  113 
thus "\<forall>(x::int).(\<forall>j\<in>{1 .. D}. \<forall>b\<in>B. x \<noteq> b + j)\<longrightarrow> (d dvd x+t) \<longrightarrow> (d dvd (x  D) + t)" by simp 
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next 

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assume d: "d dvd D" 

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{fix x assume H: "\<not>(d dvd x + t)" with d have "\<not> d dvd (x  D) + t" 
29667  117 
by (clarsimp simp add: dvd_def,erule_tac x= "ka + k" in allE,simp add: algebra_simps)} 
23465  118 
thus "\<forall>(x::int).(\<forall>j\<in>{1 .. D}. \<forall>b\<in>B. x \<noteq> b + j)\<longrightarrow> (\<not>d dvd x+t) \<longrightarrow> (\<not>d dvd (x  D) + t)" by auto 
119 
qed blast 

120 

121 
lemma aset: 

122 
"\<lbrakk>\<forall>x.(\<forall>j\<in>{1 .. D}. \<forall>b\<in>A. x \<noteq> b  j)\<longrightarrow> P x \<longrightarrow> P(x + D) ; 

123 
\<forall>x.(\<forall>j\<in>{1 .. D}. \<forall>b\<in>A. x \<noteq> b  j)\<longrightarrow> Q x \<longrightarrow> Q(x + D)\<rbrakk> \<Longrightarrow> 

124 
\<forall>x.(\<forall>j\<in>{1 .. D}. \<forall>b\<in>A. x \<noteq> b  j) \<longrightarrow> (P x \<and> Q x) \<longrightarrow> (P(x + D) \<and> Q (x + D))" 

125 
"\<lbrakk>\<forall>x.(\<forall>j\<in>{1 .. D}. \<forall>b\<in>A. x \<noteq> b  j)\<longrightarrow> P x \<longrightarrow> P(x + D) ; 

126 
\<forall>x.(\<forall>j\<in>{1 .. D}. \<forall>b\<in>A. x \<noteq> b  j)\<longrightarrow> Q x \<longrightarrow> Q(x + D)\<rbrakk> \<Longrightarrow> 

127 
\<forall>x.(\<forall>j\<in>{1 .. D}. \<forall>b\<in>A. x \<noteq> b  j)\<longrightarrow> (P x \<or> Q x) \<longrightarrow> (P(x + D) \<or> Q (x + D))" 

128 
"\<lbrakk>D>0; t + 1\<in> A\<rbrakk> \<Longrightarrow> (\<forall>x.(\<forall>j\<in>{1 .. D}. \<forall>b\<in>A. x \<noteq> b  j)\<longrightarrow> (x = t) \<longrightarrow> (x + D = t))" 

129 
"\<lbrakk>D>0 ; t \<in> A\<rbrakk> \<Longrightarrow>(\<forall>(x::int).(\<forall>j\<in>{1 .. D}. \<forall>b\<in>A. x \<noteq> b  j)\<longrightarrow> (x \<noteq> t) \<longrightarrow> (x + D \<noteq> t))" 

130 
"\<lbrakk>D>0; t\<in> A\<rbrakk> \<Longrightarrow>(\<forall>(x::int). (\<forall>j\<in>{1 .. D}. \<forall>b\<in>A. x \<noteq> b  j)\<longrightarrow> (x < t) \<longrightarrow> (x + D < t))" 

131 
"\<lbrakk>D>0; t + 1 \<in> A\<rbrakk> \<Longrightarrow> (\<forall>(x::int).(\<forall>j\<in>{1 .. D}. \<forall>b\<in>A. x \<noteq> b  j)\<longrightarrow> (x \<le> t) \<longrightarrow> (x + D \<le> t))" 

132 
"D>0 \<Longrightarrow>(\<forall>(x::int).(\<forall>j\<in>{1 .. D}. \<forall>b\<in>A. x \<noteq> b  j)\<longrightarrow> (x > t) \<longrightarrow> (x + D > t))" 

133 
"D>0 \<Longrightarrow>(\<forall>(x::int).(\<forall>j\<in>{1 .. D}. \<forall>b\<in>A. x \<noteq> b  j)\<longrightarrow> (x \<ge> t) \<longrightarrow> (x + D \<ge> t))" 

134 
"d dvd D \<Longrightarrow>(\<forall>(x::int).(\<forall>j\<in>{1 .. D}. \<forall>b\<in>A. x \<noteq> b  j)\<longrightarrow> (d dvd x+t) \<longrightarrow> (d dvd (x + D) + t))" 

135 
"d dvd D \<Longrightarrow>(\<forall>(x::int).(\<forall>j\<in>{1 .. D}. \<forall>b\<in>A. x \<noteq> b  j)\<longrightarrow> (\<not>d dvd x+t) \<longrightarrow> (\<not> d dvd (x + D) + t))" 

136 
"\<forall>x.(\<forall>j\<in>{1 .. D}. \<forall>b\<in>A. x \<noteq> b  j) \<longrightarrow> F \<longrightarrow> F" 

137 
proof (blast, blast) 

138 
assume dp: "D > 0" and tA: "t + 1 \<in> A" 

139 
show "(\<forall>x.(\<forall>j\<in>{1 .. D}. \<forall>b\<in>A. x \<noteq> b  j)\<longrightarrow> (x = t) \<longrightarrow> (x + D = t))" 

140 
apply (rule allI, rule impI,erule ballE[where x="1"],erule ballE[where x="t + 1"]) 

141 
using dp tA by simp_all 

142 
next 

143 
assume dp: "D > 0" and tA: "t \<in> A" 

144 
show "(\<forall>x.(\<forall>j\<in>{1 .. D}. \<forall>b\<in>A. x \<noteq> b  j)\<longrightarrow> (x \<noteq> t) \<longrightarrow> (x + D \<noteq> t))" 

145 
apply (rule allI, rule impI,erule ballE[where x="D"],erule ballE[where x="t"]) 

146 
using dp tA by simp_all 

147 
next 

148 
assume dp: "D > 0" thus "(\<forall>x.(\<forall>j\<in>{1 .. D}. \<forall>b\<in>A. x \<noteq> b  j)\<longrightarrow> (x > t) \<longrightarrow> (x + D > t))" by arith 

149 
next 

150 
assume dp: "D > 0" thus "\<forall>x.(\<forall>j\<in>{1 .. D}. \<forall>b\<in>A. x \<noteq> b  j)\<longrightarrow> (x \<ge> t) \<longrightarrow> (x + D \<ge> t)" by arith 

151 
next 

152 
assume dp: "D > 0" and tA:"t \<in> A" 

153 
{fix x assume nob: "\<forall>j\<in>{1 .. D}. \<forall>b\<in>A. x \<noteq> b  j" and g: "x < t" and ng: "\<not> (x + D) < t" 

154 
hence "t  x \<le> D" and "1 \<le> t  x" by simp+ 

155 
hence "\<exists>j \<in> {1 .. D}. t  x = j" by auto 

29667  156 
hence "\<exists>j \<in> {1 .. D}. x = t  j" by (auto simp add: algebra_simps) 
23465  157 
with nob tA have "False" by simp} 
158 
thus "\<forall>x.(\<forall>j\<in>{1 .. D}. \<forall>b\<in>A. x \<noteq> b  j)\<longrightarrow> (x < t) \<longrightarrow> (x + D < t)" by blast 

159 
next 

160 
assume dp: "D > 0" and tA:"t + 1\<in> A" 

161 
{fix x assume nob: "\<forall>j\<in>{1 .. D}. \<forall>b\<in>A. x \<noteq> b  j" and g: "x \<le> t" and ng: "\<not> (x + D) \<le> t" 

29667  162 
hence "(t + 1)  x \<le> D" and "1 \<le> (t + 1)  x" by (simp_all add: algebra_simps) 
23465  163 
hence "\<exists>j \<in> {1 .. D}. (t + 1)  x = j" by auto 
29667  164 
hence "\<exists>j \<in> {1 .. D}. x = (t + 1)  j" by (auto simp add: algebra_simps) 
23465  165 
with nob tA have "False" by simp} 
166 
thus "\<forall>x.(\<forall>j\<in>{1 .. D}. \<forall>b\<in>A. x \<noteq> b  j)\<longrightarrow> (x \<le> t) \<longrightarrow> (x + D \<le> t)" by blast 

167 
next 

168 
assume d: "d dvd D" 

169 
{fix x assume H: "d dvd x + t" with d have "d dvd (x + D) + t" 

29667  170 
by (clarsimp simp add: dvd_def,rule_tac x= "ka + k" in exI,simp add: algebra_simps)} 
23465  171 
thus "\<forall>(x::int).(\<forall>j\<in>{1 .. D}. \<forall>b\<in>A. x \<noteq> b  j)\<longrightarrow> (d dvd x+t) \<longrightarrow> (d dvd (x + D) + t)" by simp 
172 
next 

173 
assume d: "d dvd D" 

174 
{fix x assume H: "\<not>(d dvd x + t)" with d have "\<not>d dvd (x + D) + t" 

29667  175 
by (clarsimp simp add: dvd_def,erule_tac x= "ka  k" in allE,simp add: algebra_simps)} 
23465  176 
thus "\<forall>(x::int).(\<forall>j\<in>{1 .. D}. \<forall>b\<in>A. x \<noteq> b  j)\<longrightarrow> (\<not>d dvd x+t) \<longrightarrow> (\<not>d dvd (x + D) + t)" by auto 
177 
qed blast 

178 

179 
subsection{* Cooper's Theorem @{text "\<infinity>"} and @{text "+\<infinity>"} Version *} 

180 

181 
subsubsection{* First some trivial facts about periodic sets or predicates *} 

182 
lemma periodic_finite_ex: 

183 
assumes dpos: "(0::int) < d" and modd: "ALL x k. P x = P(x  k*d)" 

184 
shows "(EX x. P x) = (EX j : {1..d}. P j)" 

185 
(is "?LHS = ?RHS") 

186 
proof 

187 
assume ?LHS 

188 
then obtain x where P: "P x" .. 

189 
have "x mod d = x  (x div d)*d" by(simp add:zmod_zdiv_equality mult_ac eq_diff_eq) 

190 
hence Pmod: "P x = P(x mod d)" using modd by simp 

191 
show ?RHS 

192 
proof (cases) 

193 
assume "x mod d = 0" 

194 
hence "P 0" using P Pmod by simp 

195 
moreover have "P 0 = P(0  (1)*d)" using modd by blast 

196 
ultimately have "P d" by simp 

35216  197 
moreover have "d : {1..d}" using dpos by simp 
23465  198 
ultimately show ?RHS .. 
199 
next 

200 
assume not0: "x mod d \<noteq> 0" 

35216  201 
have "P(x mod d)" using dpos P Pmod by simp 
23465  202 
moreover have "x mod d : {1..d}" 
203 
proof  

204 
from dpos have "0 \<le> x mod d" by(rule pos_mod_sign) 

205 
moreover from dpos have "x mod d < d" by(rule pos_mod_bound) 

35216  206 
ultimately show ?thesis using not0 by simp 
23465  207 
qed 
208 
ultimately show ?RHS .. 

209 
qed 

210 
qed auto 

211 

212 
subsubsection{* The @{text "\<infinity>"} Version*} 

213 

214 
lemma decr_lemma: "0 < (d::int) \<Longrightarrow> x  (abs(xz)+1) * d < z" 

215 
by(induct rule: int_gr_induct,simp_all add:int_distrib) 

216 

217 
lemma incr_lemma: "0 < (d::int) \<Longrightarrow> z < x + (abs(xz)+1) * d" 

218 
by(induct rule: int_gr_induct, simp_all add:int_distrib) 

219 

220 
lemma decr_mult_lemma: 

221 
assumes dpos: "(0::int) < d" and minus: "\<forall>x. P x \<longrightarrow> P(x  d)" and knneg: "0 <= k" 

222 
shows "ALL x. P x \<longrightarrow> P(x  k*d)" 

223 
using knneg 

224 
proof (induct rule:int_ge_induct) 

225 
case base thus ?case by simp 

226 
next 

227 
case (step i) 

228 
{fix x 

229 
have "P x \<longrightarrow> P (x  i * d)" using step.hyps by blast 

230 
also have "\<dots> \<longrightarrow> P(x  (i + 1) * d)" using minus[THEN spec, of "x  i * d"] 

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by (simp add: algebra_simps) 
23465  232 
ultimately have "P x \<longrightarrow> P(x  (i + 1) * d)" by blast} 
233 
thus ?case .. 

234 
qed 

235 

236 
lemma minusinfinity: 

237 
assumes dpos: "0 < d" and 

238 
P1eqP1: "ALL x k. P1 x = P1(x  k*d)" and ePeqP1: "EX z::int. ALL x. x < z \<longrightarrow> (P x = P1 x)" 

239 
shows "(EX x. P1 x) \<longrightarrow> (EX x. P x)" 

240 
proof 

241 
assume eP1: "EX x. P1 x" 

242 
then obtain x where P1: "P1 x" .. 

243 
from ePeqP1 obtain z where P1eqP: "ALL x. x < z \<longrightarrow> (P x = P1 x)" .. 

244 
let ?w = "x  (abs(xz)+1) * d" 

245 
from dpos have w: "?w < z" by(rule decr_lemma) 

246 
have "P1 x = P1 ?w" using P1eqP1 by blast 

247 
also have "\<dots> = P(?w)" using w P1eqP by blast 

248 
finally have "P ?w" using P1 by blast 

249 
thus "EX x. P x" .. 

250 
qed 

251 

252 
lemma cpmi: 

253 
assumes dp: "0 < D" and p1:"\<exists>z. \<forall> x< z. P x = P' x" 

254 
and nb:"\<forall>x.(\<forall> j\<in> {1..D}. \<forall>(b::int) \<in> B. x \<noteq> b+j) > P (x) > P (x  D)" 

255 
and pd: "\<forall> x k. P' x = P' (xk*D)" 

256 
shows "(\<exists>x. P x) = ((\<exists> j\<in> {1..D} . P' j)  (\<exists> j \<in> {1..D}.\<exists> b\<in> B. P (b+j)))" 

257 
(is "?L = (?R1 \<or> ?R2)") 

258 
proof 

259 
{assume "?R2" hence "?L" by blast} 

260 
moreover 

261 
{assume H:"?R1" hence "?L" using minusinfinity[OF dp pd p1] periodic_finite_ex[OF dp pd] by simp} 

262 
moreover 

263 
{ fix x 

264 
assume P: "P x" and H: "\<not> ?R2" 

265 
{fix y assume "\<not> (\<exists>j\<in>{1..D}. \<exists>b\<in>B. P (b + j))" and P: "P y" 

266 
hence "~(EX (j::int) : {1..D}. EX (b::int) : B. y = b+j)" by auto 

267 
with nb P have "P (y  D)" by auto } 

268 
hence "ALL x.~(EX (j::int) : {1..D}. EX (b::int) : B. P(b+j)) > P (x) > P (x  D)" by blast 

269 
with H P have th: " \<forall>x. P x \<longrightarrow> P (x  D)" by auto 

270 
from p1 obtain z where z: "ALL x. x < z > (P x = P' x)" by blast 

271 
let ?y = "x  (\<bar>x  z\<bar> + 1)*D" 

272 
have zp: "0 <= (\<bar>x  z\<bar> + 1)" by arith 

273 
from dp have yz: "?y < z" using decr_lemma[OF dp] by simp 

274 
from z[rule_format, OF yz] decr_mult_lemma[OF dp th zp, rule_format, OF P] have th2: " P' ?y" by auto 

275 
with periodic_finite_ex[OF dp pd] 

276 
have "?R1" by blast} 

277 
ultimately show ?thesis by blast 

278 
qed 

279 

280 
subsubsection {* The @{text "+\<infinity>"} Version*} 

281 

282 
lemma plusinfinity: 

283 
assumes dpos: "(0::int) < d" and 

284 
P1eqP1: "\<forall>x k. P' x = P'(x  k*d)" and ePeqP1: "\<exists> z. \<forall> x>z. P x = P' x" 

285 
shows "(\<exists> x. P' x) \<longrightarrow> (\<exists> x. P x)" 

286 
proof 

287 
assume eP1: "EX x. P' x" 

288 
then obtain x where P1: "P' x" .. 

289 
from ePeqP1 obtain z where P1eqP: "\<forall>x>z. P x = P' x" .. 

290 
let ?w' = "x + (abs(xz)+1) * d" 

291 
let ?w = "x  ((abs(xz) + 1))*d" 

29667  292 
have ww'[simp]: "?w = ?w'" by (simp add: algebra_simps) 
23465  293 
from dpos have w: "?w > z" by(simp only: ww' incr_lemma) 
294 
hence "P' x = P' ?w" using P1eqP1 by blast 

295 
also have "\<dots> = P(?w)" using w P1eqP by blast 

296 
finally have "P ?w" using P1 by blast 

297 
thus "EX x. P x" .. 

298 
qed 

299 

300 
lemma incr_mult_lemma: 

301 
assumes dpos: "(0::int) < d" and plus: "ALL x::int. P x \<longrightarrow> P(x + d)" and knneg: "0 <= k" 

302 
shows "ALL x. P x \<longrightarrow> P(x + k*d)" 

303 
using knneg 

304 
proof (induct rule:int_ge_induct) 

305 
case base thus ?case by simp 

306 
next 

307 
case (step i) 

308 
{fix x 

309 
have "P x \<longrightarrow> P (x + i * d)" using step.hyps by blast 

310 
also have "\<dots> \<longrightarrow> P(x + (i + 1) * d)" using plus[THEN spec, of "x + i * d"] 

44766  311 
by (simp add:int_distrib add_ac) 
23465  312 
ultimately have "P x \<longrightarrow> P(x + (i + 1) * d)" by blast} 
313 
thus ?case .. 

314 
qed 

315 

316 
lemma cppi: 

317 
assumes dp: "0 < D" and p1:"\<exists>z. \<forall> x> z. P x = P' x" 

318 
and nb:"\<forall>x.(\<forall> j\<in> {1..D}. \<forall>(b::int) \<in> A. x \<noteq> b  j) > P (x) > P (x + D)" 

319 
and pd: "\<forall> x k. P' x= P' (xk*D)" 

320 
shows "(\<exists>x. P x) = ((\<exists> j\<in> {1..D} . P' j)  (\<exists> j \<in> {1..D}.\<exists> b\<in> A. P (b  j)))" (is "?L = (?R1 \<or> ?R2)") 

321 
proof 

322 
{assume "?R2" hence "?L" by blast} 

323 
moreover 

324 
{assume H:"?R1" hence "?L" using plusinfinity[OF dp pd p1] periodic_finite_ex[OF dp pd] by simp} 

325 
moreover 

326 
{ fix x 

327 
assume P: "P x" and H: "\<not> ?R2" 

328 
{fix y assume "\<not> (\<exists>j\<in>{1..D}. \<exists>b\<in>A. P (b  j))" and P: "P y" 

329 
hence "~(EX (j::int) : {1..D}. EX (b::int) : A. y = b  j)" by auto 

330 
with nb P have "P (y + D)" by auto } 

331 
hence "ALL x.~(EX (j::int) : {1..D}. EX (b::int) : A. P(bj)) > P (x) > P (x + D)" by blast 

332 
with H P have th: " \<forall>x. P x \<longrightarrow> P (x + D)" by auto 

333 
from p1 obtain z where z: "ALL x. x > z > (P x = P' x)" by blast 

334 
let ?y = "x + (\<bar>x  z\<bar> + 1)*D" 

335 
have zp: "0 <= (\<bar>x  z\<bar> + 1)" by arith 

336 
from dp have yz: "?y > z" using incr_lemma[OF dp] by simp 

337 
from z[rule_format, OF yz] incr_mult_lemma[OF dp th zp, rule_format, OF P] have th2: " P' ?y" by auto 

338 
with periodic_finite_ex[OF dp pd] 

339 
have "?R1" by blast} 

340 
ultimately show ?thesis by blast 

341 
qed 

342 

343 
lemma simp_from_to: "{i..j::int} = (if j < i then {} else insert i {i+1..j})" 

344 
apply(simp add:atLeastAtMost_def atLeast_def atMost_def) 

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apply(fastforce) 
23465  346 
done 
347 

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theorem unity_coeff_ex: "(\<exists>(x::'a::{semiring_0,Rings.dvd}). P (l * x)) \<equiv> (\<exists>x. l dvd (x + 0) \<and> P x)" 
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apply (rule eq_reflection [symmetric]) 
23465  350 
apply (rule iffI) 
351 
defer 

352 
apply (erule exE) 

353 
apply (rule_tac x = "l * x" in exI) 

354 
apply (simp add: dvd_def) 

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apply (rule_tac x = x in exI, simp) 
23465  356 
apply (erule exE) 
357 
apply (erule conjE) 

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358 
apply simp 
23465  359 
apply (erule dvdE) 
360 
apply (rule_tac x = k in exI) 

361 
apply simp 

362 
done 

363 

364 
lemma zdvd_mono: assumes not0: "(k::int) \<noteq> 0" 

365 
shows "((m::int) dvd t) \<equiv> (k*m dvd k*t)" 

366 
using not0 by (simp add: dvd_def) 

367 

368 
lemma uminus_dvd_conv: "(d dvd (t::int)) \<equiv> (d dvd t)" "(d dvd (t::int)) \<equiv> (d dvd t)" 

369 
by simp_all 

32553  370 

23465  371 
text {* \bigskip Theorems for transforming predicates on nat to predicates on @{text int}*} 
32553  372 

23465  373 
lemma zdiff_int_split: "P (int (x  y)) = 
374 
((y \<le> x \<longrightarrow> P (int x  int y)) \<and> (x < y \<longrightarrow> P 0))" 

36800  375 
by (cases "y \<le> x") (simp_all add: zdiff_int) 
23465  376 

377 
text {* 

378 
\medskip Specific instances of congruence rules, to prevent 

379 
simplifier from looping. *} 

380 

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theorem imp_le_cong: 
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382 
"\<lbrakk>x = x'; 0 \<le> x' \<Longrightarrow> P = P'\<rbrakk> \<Longrightarrow> (0 \<le> (x::int) \<longrightarrow> P) = (0 \<le> x' \<longrightarrow> P')" 
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383 
by simp 
23465  384 

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theorem conj_le_cong: 
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386 
"\<lbrakk>x = x'; 0 \<le> x' \<Longrightarrow> P = P'\<rbrakk> \<Longrightarrow> (0 \<le> (x::int) \<and> P) = (0 \<le> x' \<and> P')" 
23465  387 
by (simp cong: conj_cong) 
36799  388 

36798  389 
use "Tools/Qelim/cooper.ML" 
23465  390 

36799  391 
setup Cooper.setup 
23465  392 

36804  393 
method_setup presburger = "Cooper.method" "Cooper's algorithm for Presburger arithmetic" 
23465  394 

36798  395 
declare dvd_eq_mod_eq_0[symmetric, presburger] 
396 
declare mod_1[presburger] 

397 
declare mod_0[presburger] 

398 
declare mod_by_1[presburger] 

399 
declare mod_self[presburger] 

400 
declare mod_by_0[presburger] 

401 
declare mod_div_trivial[presburger] 

402 
declare div_mod_equality2[presburger] 

403 
declare div_mod_equality[presburger] 

404 
declare mod_div_equality2[presburger] 

405 
declare mod_div_equality[presburger] 

406 
declare mod_mult_self1[presburger] 

407 
declare mod_mult_self2[presburger] 

47165  408 
declare div_mod_equality[presburger] 
409 
declare div_mod_equality2[presburger] 

36798  410 
declare mod2_Suc_Suc[presburger] 
411 
lemma [presburger]: "(a::int) div 0 = 0" and [presburger]: "a mod 0 = a" 

412 
by simp_all 

413 

27668  414 
lemma [presburger, algebra]: "m mod 2 = (1::nat) \<longleftrightarrow> \<not> 2 dvd m " by presburger 
415 
lemma [presburger, algebra]: "m mod 2 = Suc 0 \<longleftrightarrow> \<not> 2 dvd m " by presburger 

416 
lemma [presburger, algebra]: "m mod (Suc (Suc 0)) = (1::nat) \<longleftrightarrow> \<not> 2 dvd m " by presburger 

417 
lemma [presburger, algebra]: "m mod (Suc (Suc 0)) = Suc 0 \<longleftrightarrow> \<not> 2 dvd m " by presburger 

418 
lemma [presburger, algebra]: "m mod 2 = (1::int) \<longleftrightarrow> \<not> 2 dvd m " by presburger 

23465  419 

420 
end 