(* Title: HOL/Library/Boolean_Algebra.thy
Author: Brian Huffman
*)
header {* Boolean Algebras *}
theory Boolean_Algebra
imports Plain
begin
locale boolean =
fixes conj :: "'a \<Rightarrow> 'a \<Rightarrow> 'a" (infixr "\<sqinter>" 70)
fixes disj :: "'a \<Rightarrow> 'a \<Rightarrow> 'a" (infixr "\<squnion>" 65)
fixes compl :: "'a \<Rightarrow> 'a" ("\<sim> _" [81] 80)
fixes zero :: "'a" ("\<zero>")
fixes one :: "'a" ("\<one>")
assumes conj_assoc: "(x \<sqinter> y) \<sqinter> z = x \<sqinter> (y \<sqinter> z)"
assumes disj_assoc: "(x \<squnion> y) \<squnion> z = x \<squnion> (y \<squnion> z)"
assumes conj_commute: "x \<sqinter> y = y \<sqinter> x"
assumes disj_commute: "x \<squnion> y = y \<squnion> x"
assumes conj_disj_distrib: "x \<sqinter> (y \<squnion> z) = (x \<sqinter> y) \<squnion> (x \<sqinter> z)"
assumes disj_conj_distrib: "x \<squnion> (y \<sqinter> z) = (x \<squnion> y) \<sqinter> (x \<squnion> z)"
assumes conj_one_right [simp]: "x \<sqinter> \<one> = x"
assumes disj_zero_right [simp]: "x \<squnion> \<zero> = x"
assumes conj_cancel_right [simp]: "x \<sqinter> \<sim> x = \<zero>"
assumes disj_cancel_right [simp]: "x \<squnion> \<sim> x = \<one>"
begin
lemmas disj_ac =
disj_assoc disj_commute
mk_left_commute [where 'a = 'a, of "disj", OF disj_assoc disj_commute]
lemmas conj_ac =
conj_assoc conj_commute
mk_left_commute [where 'a = 'a, of "conj", OF conj_assoc conj_commute]
lemma dual: "boolean disj conj compl one zero"
apply (rule boolean.intro)
apply (rule disj_assoc)
apply (rule conj_assoc)
apply (rule disj_commute)
apply (rule conj_commute)
apply (rule disj_conj_distrib)
apply (rule conj_disj_distrib)
apply (rule disj_zero_right)
apply (rule conj_one_right)
apply (rule disj_cancel_right)
apply (rule conj_cancel_right)
done
subsection {* Complement *}
lemma complement_unique:
assumes 1: "a \<sqinter> x = \<zero>"
assumes 2: "a \<squnion> x = \<one>"
assumes 3: "a \<sqinter> y = \<zero>"
assumes 4: "a \<squnion> y = \<one>"
shows "x = y"
proof -
have "(a \<sqinter> x) \<squnion> (x \<sqinter> y) = (a \<sqinter> y) \<squnion> (x \<sqinter> y)" using 1 3 by simp
hence "(x \<sqinter> a) \<squnion> (x \<sqinter> y) = (y \<sqinter> a) \<squnion> (y \<sqinter> x)" using conj_commute by simp
hence "x \<sqinter> (a \<squnion> y) = y \<sqinter> (a \<squnion> x)" using conj_disj_distrib by simp
hence "x \<sqinter> \<one> = y \<sqinter> \<one>" using 2 4 by simp
thus "x = y" using conj_one_right by simp
qed
lemma compl_unique: "\<lbrakk>x \<sqinter> y = \<zero>; x \<squnion> y = \<one>\<rbrakk> \<Longrightarrow> \<sim> x = y"
by (rule complement_unique [OF conj_cancel_right disj_cancel_right])
lemma double_compl [simp]: "\<sim> (\<sim> x) = x"
proof (rule compl_unique)
from conj_cancel_right show "\<sim> x \<sqinter> x = \<zero>" by (simp only: conj_commute)
from disj_cancel_right show "\<sim> x \<squnion> x = \<one>" by (simp only: disj_commute)
qed
lemma compl_eq_compl_iff [simp]: "(\<sim> x = \<sim> y) = (x = y)"
by (rule inj_eq [OF inj_on_inverseI], rule double_compl)
subsection {* Conjunction *}
lemma conj_absorb [simp]: "x \<sqinter> x = x"
proof -
have "x \<sqinter> x = (x \<sqinter> x) \<squnion> \<zero>" using disj_zero_right by simp
also have "... = (x \<sqinter> x) \<squnion> (x \<sqinter> \<sim> x)" using conj_cancel_right by simp
also have "... = x \<sqinter> (x \<squnion> \<sim> x)" using conj_disj_distrib by (simp only:)
also have "... = x \<sqinter> \<one>" using disj_cancel_right by simp
also have "... = x" using conj_one_right by simp
finally show ?thesis .
qed
lemma conj_zero_right [simp]: "x \<sqinter> \<zero> = \<zero>"
proof -
have "x \<sqinter> \<zero> = x \<sqinter> (x \<sqinter> \<sim> x)" using conj_cancel_right by simp
also have "... = (x \<sqinter> x) \<sqinter> \<sim> x" using conj_assoc by (simp only:)
also have "... = x \<sqinter> \<sim> x" using conj_absorb by simp
also have "... = \<zero>" using conj_cancel_right by simp
finally show ?thesis .
qed
lemma compl_one [simp]: "\<sim> \<one> = \<zero>"
by (rule compl_unique [OF conj_zero_right disj_zero_right])
lemma conj_zero_left [simp]: "\<zero> \<sqinter> x = \<zero>"
by (subst conj_commute) (rule conj_zero_right)
lemma conj_one_left [simp]: "\<one> \<sqinter> x = x"
by (subst conj_commute) (rule conj_one_right)
lemma conj_cancel_left [simp]: "\<sim> x \<sqinter> x = \<zero>"
by (subst conj_commute) (rule conj_cancel_right)
lemma conj_left_absorb [simp]: "x \<sqinter> (x \<sqinter> y) = x \<sqinter> y"
by (simp only: conj_assoc [symmetric] conj_absorb)
lemma conj_disj_distrib2:
"(y \<squnion> z) \<sqinter> x = (y \<sqinter> x) \<squnion> (z \<sqinter> x)"
by (simp only: conj_commute conj_disj_distrib)
lemmas conj_disj_distribs =
conj_disj_distrib conj_disj_distrib2
subsection {* Disjunction *}
lemma disj_absorb [simp]: "x \<squnion> x = x"
by (rule boolean.conj_absorb [OF dual])
lemma disj_one_right [simp]: "x \<squnion> \<one> = \<one>"
by (rule boolean.conj_zero_right [OF dual])
lemma compl_zero [simp]: "\<sim> \<zero> = \<one>"
by (rule boolean.compl_one [OF dual])
lemma disj_zero_left [simp]: "\<zero> \<squnion> x = x"
by (rule boolean.conj_one_left [OF dual])
lemma disj_one_left [simp]: "\<one> \<squnion> x = \<one>"
by (rule boolean.conj_zero_left [OF dual])
lemma disj_cancel_left [simp]: "\<sim> x \<squnion> x = \<one>"
by (rule boolean.conj_cancel_left [OF dual])
lemma disj_left_absorb [simp]: "x \<squnion> (x \<squnion> y) = x \<squnion> y"
by (rule boolean.conj_left_absorb [OF dual])
lemma disj_conj_distrib2:
"(y \<sqinter> z) \<squnion> x = (y \<squnion> x) \<sqinter> (z \<squnion> x)"
by (rule boolean.conj_disj_distrib2 [OF dual])
lemmas disj_conj_distribs =
disj_conj_distrib disj_conj_distrib2
subsection {* De Morgan's Laws *}
lemma de_Morgan_conj [simp]: "\<sim> (x \<sqinter> y) = \<sim> x \<squnion> \<sim> y"
proof (rule compl_unique)
have "(x \<sqinter> y) \<sqinter> (\<sim> x \<squnion> \<sim> y) = ((x \<sqinter> y) \<sqinter> \<sim> x) \<squnion> ((x \<sqinter> y) \<sqinter> \<sim> y)"
by (rule conj_disj_distrib)
also have "... = (y \<sqinter> (x \<sqinter> \<sim> x)) \<squnion> (x \<sqinter> (y \<sqinter> \<sim> y))"
by (simp only: conj_ac)
finally show "(x \<sqinter> y) \<sqinter> (\<sim> x \<squnion> \<sim> y) = \<zero>"
by (simp only: conj_cancel_right conj_zero_right disj_zero_right)
next
have "(x \<sqinter> y) \<squnion> (\<sim> x \<squnion> \<sim> y) = (x \<squnion> (\<sim> x \<squnion> \<sim> y)) \<sqinter> (y \<squnion> (\<sim> x \<squnion> \<sim> y))"
by (rule disj_conj_distrib2)
also have "... = (\<sim> y \<squnion> (x \<squnion> \<sim> x)) \<sqinter> (\<sim> x \<squnion> (y \<squnion> \<sim> y))"
by (simp only: disj_ac)
finally show "(x \<sqinter> y) \<squnion> (\<sim> x \<squnion> \<sim> y) = \<one>"
by (simp only: disj_cancel_right disj_one_right conj_one_right)
qed
lemma de_Morgan_disj [simp]: "\<sim> (x \<squnion> y) = \<sim> x \<sqinter> \<sim> y"
by (rule boolean.de_Morgan_conj [OF dual])
end
subsection {* Symmetric Difference *}
locale boolean_xor = boolean +
fixes xor :: "'a => 'a => 'a" (infixr "\<oplus>" 65)
assumes xor_def: "x \<oplus> y = (x \<sqinter> \<sim> y) \<squnion> (\<sim> x \<sqinter> y)"
begin
lemma xor_def2:
"x \<oplus> y = (x \<squnion> y) \<sqinter> (\<sim> x \<squnion> \<sim> y)"
by (simp only: xor_def conj_disj_distribs
disj_ac conj_ac conj_cancel_right disj_zero_left)
lemma xor_commute: "x \<oplus> y = y \<oplus> x"
by (simp only: xor_def conj_commute disj_commute)
lemma xor_assoc: "(x \<oplus> y) \<oplus> z = x \<oplus> (y \<oplus> z)"
proof -
let ?t = "(x \<sqinter> y \<sqinter> z) \<squnion> (x \<sqinter> \<sim> y \<sqinter> \<sim> z) \<squnion>
(\<sim> x \<sqinter> y \<sqinter> \<sim> z) \<squnion> (\<sim> x \<sqinter> \<sim> y \<sqinter> z)"
have "?t \<squnion> (z \<sqinter> x \<sqinter> \<sim> x) \<squnion> (z \<sqinter> y \<sqinter> \<sim> y) =
?t \<squnion> (x \<sqinter> y \<sqinter> \<sim> y) \<squnion> (x \<sqinter> z \<sqinter> \<sim> z)"
by (simp only: conj_cancel_right conj_zero_right)
thus "(x \<oplus> y) \<oplus> z = x \<oplus> (y \<oplus> z)"
apply (simp only: xor_def de_Morgan_disj de_Morgan_conj double_compl)
apply (simp only: conj_disj_distribs conj_ac disj_ac)
done
qed
lemmas xor_ac =
xor_assoc xor_commute
mk_left_commute [where 'a = 'a, of "xor", OF xor_assoc xor_commute]
lemma xor_zero_right [simp]: "x \<oplus> \<zero> = x"
by (simp only: xor_def compl_zero conj_one_right conj_zero_right disj_zero_right)
lemma xor_zero_left [simp]: "\<zero> \<oplus> x = x"
by (subst xor_commute) (rule xor_zero_right)
lemma xor_one_right [simp]: "x \<oplus> \<one> = \<sim> x"
by (simp only: xor_def compl_one conj_zero_right conj_one_right disj_zero_left)
lemma xor_one_left [simp]: "\<one> \<oplus> x = \<sim> x"
by (subst xor_commute) (rule xor_one_right)
lemma xor_self [simp]: "x \<oplus> x = \<zero>"
by (simp only: xor_def conj_cancel_right conj_cancel_left disj_zero_right)
lemma xor_left_self [simp]: "x \<oplus> (x \<oplus> y) = y"
by (simp only: xor_assoc [symmetric] xor_self xor_zero_left)
lemma xor_compl_left: "\<sim> x \<oplus> y = \<sim> (x \<oplus> y)"
apply (simp only: xor_def de_Morgan_disj de_Morgan_conj double_compl)
apply (simp only: conj_disj_distribs)
apply (simp only: conj_cancel_right conj_cancel_left)
apply (simp only: disj_zero_left disj_zero_right)
apply (simp only: disj_ac conj_ac)
done
lemma xor_compl_right: "x \<oplus> \<sim> y = \<sim> (x \<oplus> y)"
apply (simp only: xor_def de_Morgan_disj de_Morgan_conj double_compl)
apply (simp only: conj_disj_distribs)
apply (simp only: conj_cancel_right conj_cancel_left)
apply (simp only: disj_zero_left disj_zero_right)
apply (simp only: disj_ac conj_ac)
done
lemma xor_cancel_right [simp]: "x \<oplus> \<sim> x = \<one>"
by (simp only: xor_compl_right xor_self compl_zero)
lemma xor_cancel_left [simp]: "\<sim> x \<oplus> x = \<one>"
by (subst xor_commute) (rule xor_cancel_right)
lemma conj_xor_distrib: "x \<sqinter> (y \<oplus> z) = (x \<sqinter> y) \<oplus> (x \<sqinter> z)"
proof -
have "(x \<sqinter> y \<sqinter> \<sim> z) \<squnion> (x \<sqinter> \<sim> y \<sqinter> z) =
(y \<sqinter> x \<sqinter> \<sim> x) \<squnion> (z \<sqinter> x \<sqinter> \<sim> x) \<squnion> (x \<sqinter> y \<sqinter> \<sim> z) \<squnion> (x \<sqinter> \<sim> y \<sqinter> z)"
by (simp only: conj_cancel_right conj_zero_right disj_zero_left)
thus "x \<sqinter> (y \<oplus> z) = (x \<sqinter> y) \<oplus> (x \<sqinter> z)"
by (simp (no_asm_use) only:
xor_def de_Morgan_disj de_Morgan_conj double_compl
conj_disj_distribs conj_ac disj_ac)
qed
lemma conj_xor_distrib2:
"(y \<oplus> z) \<sqinter> x = (y \<sqinter> x) \<oplus> (z \<sqinter> x)"
proof -
have "x \<sqinter> (y \<oplus> z) = (x \<sqinter> y) \<oplus> (x \<sqinter> z)"
by (rule conj_xor_distrib)
thus "(y \<oplus> z) \<sqinter> x = (y \<sqinter> x) \<oplus> (z \<sqinter> x)"
by (simp only: conj_commute)
qed
lemmas conj_xor_distribs =
conj_xor_distrib conj_xor_distrib2
end
end