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section "Some Auxiliary Results"
theory Auxiliary imports Main
begin
lemma disjE3: "P \<or> Q \<or> R \<Longrightarrow> (P \<Longrightarrow> S) \<Longrightarrow> (Q \<Longrightarrow> S) \<Longrightarrow> (R \<Longrightarrow> S) \<Longrightarrow> S" by auto
lemma ge_induct[consumes 1, case_names step]:
fixes i::nat and j::nat and P::"nat \<Rightarrow> bool"
shows "i \<le> j \<Longrightarrow> (\<And>n. i \<le> n \<Longrightarrow> ((\<forall>m \<ge> i. m<n \<longrightarrow> P m) \<Longrightarrow> P n)) \<Longrightarrow> P j"
proof -
assume a0: "i \<le> j" and a1: "(\<And>n. i \<le> n \<Longrightarrow> ((\<forall>m \<ge> i. m<n \<longrightarrow> P m) \<Longrightarrow> P n))"
have "(\<And>n. \<forall>m<n. i \<le> m \<longrightarrow> P m \<Longrightarrow> i \<le> n \<longrightarrow> P n)"
proof
fix n
assume a2: "\<forall>m<n. i \<le> m \<longrightarrow> P m"
show "i \<le> n \<Longrightarrow> P n"
proof -
assume "i \<le> n"
with a1 have "(\<forall>m \<ge> i. m<n \<longrightarrow> P m) \<Longrightarrow> P n" by simp
moreover from a2 have "\<forall>m \<ge> i. m<n \<longrightarrow> P m" by simp
ultimately show "P n" by simp
qed
qed
with nat_less_induct[of "\<lambda>j. i \<le> j \<longrightarrow> P j" j] have "i \<le> j \<longrightarrow> P j" .
with a0 show ?thesis by simp
qed
lemma my_induct[consumes 1, case_names base step]:
fixes P::"nat\<Rightarrow>bool"
assumes less: "i \<le> j"
and base: "P j"
and step: "\<And>n. i \<le> n \<Longrightarrow> n < j \<Longrightarrow> (\<forall>n'>n. n'\<le>j \<longrightarrow> P n') \<Longrightarrow> P n"
shows "P i"
proof cases
assume "j=0"
thus ?thesis using less base by simp
next
assume "\<not> j=0"
have "j - (j - i) \<ge> i \<longrightarrow> P (j - (j - i))"
proof (rule less_induct[of "\<lambda>n::nat. j-n \<ge> i \<longrightarrow> P (j-n)" "j-i"])
fix x assume asmp: "\<And>y. y < x \<Longrightarrow> i \<le> j - y \<longrightarrow> P (j - y)"
show "i \<le> j - x \<longrightarrow> P (j - x)"
proof cases
assume "x=0"
with base show ?thesis by simp
next
assume "\<not> x=0"
with \<open>j \<noteq> 0\<close> have "j - x < j" by simp
show ?thesis
proof
assume "i \<le> j - x"
moreover have "\<forall>n'>j-x. n'\<le>j \<longrightarrow> P n'"
proof
fix n'
show "n'>j-x \<longrightarrow> n'\<le>j \<longrightarrow> P n'"
proof (rule HOL.impI[OF HOL.impI])
assume "j - x < n'" and "n' \<le> j"
hence "j - n' < x" by simp
moreover from \<open>i \<le> j - x\<close> \<open>j - x < n'\<close> have "i \<le> n'" using le_less_trans less_imp_le_nat by blast
with \<open>n' \<le> j\<close> have "i \<le> j - (j - n')" by simp
ultimately have "P (j - (j - n'))" using asmp by simp
moreover from \<open>n' \<le> j\<close> have "j - (j - n') = n'" by simp
ultimately show "P n'" by simp
qed
qed
ultimately show "P (j - x)" using \<open>j-x<j\<close> step[of "j-x"] by simp
qed
qed
qed
moreover from less have "j - (j - i) = i" by simp
ultimately show ?thesis by simp
qed
lemma Greatest_ex_le_nat: assumes "\<exists>k. P k \<and> (\<forall>k'. P k' \<longrightarrow> k' \<le> k)" shows "\<not>(\<exists>n'>Greatest P. P n')"
by (metis Greatest_equality assms less_le_not_le)
lemma cardEx: assumes "finite A" and "finite B" and "card A > card B" shows "\<exists>x\<in>A. \<not> x\<in>B"
proof cases
assume "A \<subseteq> B"
with assms have "card A\<le>card B" using card_mono by blast
with assms have False by simp
thus ?thesis by simp
next
assume "\<not> A \<subseteq> B"
thus ?thesis by auto
qed
lemma cardshift: "card {i::nat. i>n \<and> i \<le> n' \<and> p (n'' + i)} = card {i. i>(n + n'') \<and> i \<le> (n' + n'') \<and> p i}"
proof -
let ?f="\<lambda>i. i+n''"
have "bij_betw ?f {i::nat. i>n \<and> i \<le> n' \<and> p (n'' + i)} {i. i>(n + n'') \<and> i \<le> (n' + n'') \<and> p i}"
proof (rule bij_betwI')
fix x y assume "x \<in> {i. n < i \<and> i \<le> n' \<and> p (n'' + i)}" and "y \<in> {i. n < i \<and> i \<le> n' \<and> p (n'' + i)}"
show "(x + n'' = y + n'') = (x = y)" by simp
next
fix x::nat assume "x \<in> {i. n < i \<and> i \<le> n' \<and> p (n'' + i)}"
hence "n<x" and "x \<le> n'" and "p(n''+x)" by auto
moreover have "n''+x=x+n''" by simp
ultimately have "n + n'' < x + n''" and "x + n'' \<le> n' + n''" and "p (x + n'')" by auto
thus "x + n'' \<in> {i. n + n'' < i \<and> i \<le> n' + n'' \<and> p i}" by auto
next
fix y::nat assume "y \<in> {i. n + n'' < i \<and> i \<le> n' + n'' \<and> p i}"
hence "n+n''<y" and "y\<le>n'+n''" and "p y" by auto
then obtain x where "x=y-n''" by simp
with \<open>n+n''<y\<close> have "y=x+n''" by simp
moreover from \<open>x=y-n''\<close> \<open>n+n''<y\<close> have "x>n" by simp
moreover from \<open>x=y-n''\<close> \<open>y\<le>n'+n''\<close> have "x\<le>n'" by simp
moreover from \<open>y=x+n''\<close> have "y=n''+x" by simp
with \<open>p y\<close> have "p (n'' + x)" by simp
ultimately show "\<exists>x\<in>{i. n < i \<and> i \<le> n' \<and> p (n'' + i)}. y = x + n''" by auto
qed
thus ?thesis using bij_betw_same_card by auto
qed
end |