Theory Boolean_Algebra

theory Boolean_Algebra
imports Main
(*  Title:      HOL/Library/Boolean_Algebra.thy
Author: Brian Huffman
*)


header {* Boolean Algebras *}

theory Boolean_Algebra
imports Main
begin

locale boolean =
fixes conj :: "'a => 'a => 'a" (infixr "\<sqinter>" 70)
fixes disj :: "'a => 'a => 'a" (infixr "\<squnion>" 65)
fixes compl :: "'a => 'a" ("∼ _" [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> ∼ x = \<zero>"
assumes disj_cancel_right [simp]: "x \<squnion> ∼ x = \<one>"

sublocale boolean < conj!: abel_semigroup conj proof
qed (fact conj_assoc conj_commute)+

sublocale boolean < disj!: abel_semigroup disj proof
qed (fact disj_assoc disj_commute)+

context boolean
begin

lemmas conj_left_commute = conj.left_commute

lemmas disj_left_commute = disj.left_commute

lemmas conj_ac = conj.assoc conj.commute conj.left_commute
lemmas disj_ac = disj.assoc disj.commute disj.left_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: "[|x \<sqinter> y = \<zero>; x \<squnion> y = \<one>|] ==> ∼ x = y"
by (rule complement_unique [OF conj_cancel_right disj_cancel_right])

lemma double_compl [simp]: "∼ (∼ x) = x"
proof (rule compl_unique)
from conj_cancel_right show "∼ x \<sqinter> x = \<zero>" by (simp only: conj_commute)
from disj_cancel_right show "∼ x \<squnion> x = \<one>" by (simp only: disj_commute)
qed

lemma compl_eq_compl_iff [simp]: "(∼ x = ∼ 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> ∼ x)" using conj_cancel_right by simp
also have "... = x \<sqinter> (x \<squnion> ∼ 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> ∼ x)" using conj_cancel_right by simp
also have "... = (x \<sqinter> x) \<sqinter> ∼ x" using conj_assoc by (simp only:)
also have "... = x \<sqinter> ∼ x" using conj_absorb by simp
also have "... = \<zero>" using conj_cancel_right by simp
finally show ?thesis .
qed

lemma compl_one [simp]: "∼ \<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]: "∼ 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]: "∼ \<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]: "∼ 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]: "∼ (x \<sqinter> y) = ∼ x \<squnion> ∼ y"
proof (rule compl_unique)
have "(x \<sqinter> y) \<sqinter> (∼ x \<squnion> ∼ y) = ((x \<sqinter> y) \<sqinter> ∼ x) \<squnion> ((x \<sqinter> y) \<sqinter> ∼ y)"
by (rule conj_disj_distrib)
also have "... = (y \<sqinter> (x \<sqinter> ∼ x)) \<squnion> (x \<sqinter> (y \<sqinter> ∼ y))"
by (simp only: conj_ac)
finally show "(x \<sqinter> y) \<sqinter> (∼ x \<squnion> ∼ y) = \<zero>"
by (simp only: conj_cancel_right conj_zero_right disj_zero_right)
next
have "(x \<sqinter> y) \<squnion> (∼ x \<squnion> ∼ y) = (x \<squnion> (∼ x \<squnion> ∼ y)) \<sqinter> (y \<squnion> (∼ x \<squnion> ∼ y))"
by (rule disj_conj_distrib2)
also have "... = (∼ y \<squnion> (x \<squnion> ∼ x)) \<sqinter> (∼ x \<squnion> (y \<squnion> ∼ y))"
by (simp only: disj_ac)
finally show "(x \<sqinter> y) \<squnion> (∼ x \<squnion> ∼ y) = \<one>"
by (simp only: disj_cancel_right disj_one_right conj_one_right)
qed

lemma de_Morgan_disj [simp]: "∼ (x \<squnion> y) = ∼ x \<sqinter> ∼ y"
by (rule boolean.de_Morgan_conj [OF dual])

end

subsection {* Symmetric Difference *}

locale boolean_xor = boolean +
fixes xor :: "'a => 'a => 'a" (infixr "⊕" 65)
assumes xor_def: "x ⊕ y = (x \<sqinter> ∼ y) \<squnion> (∼ x \<sqinter> y)"

sublocale boolean_xor < xor!: abel_semigroup xor proof
fix x y z :: 'a
let ?t = "(x \<sqinter> y \<sqinter> z) \<squnion> (x \<sqinter> ∼ y \<sqinter> ∼ z) \<squnion>
(∼ x \<sqinter> y \<sqinter> ∼ z) \<squnion> (∼ x \<sqinter> ∼ y \<sqinter> z)"

have "?t \<squnion> (z \<sqinter> x \<sqinter> ∼ x) \<squnion> (z \<sqinter> y \<sqinter> ∼ y) =
?t \<squnion> (x \<sqinter> y \<sqinter> ∼ y) \<squnion> (x \<sqinter> z \<sqinter> ∼ z)"

by (simp only: conj_cancel_right conj_zero_right)
thus "(x ⊕ y) ⊕ z = x ⊕ (y ⊕ 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
show "x ⊕ y = y ⊕ x"
by (simp only: xor_def conj_commute disj_commute)
qed

context boolean_xor
begin

lemmas xor_assoc = xor.assoc
lemmas xor_commute = xor.commute
lemmas xor_left_commute = xor.left_commute

lemmas xor_ac = xor.assoc xor.commute xor.left_commute

lemma xor_def2:
"x ⊕ y = (x \<squnion> y) \<sqinter> (∼ x \<squnion> ∼ y)"
by (simp only: xor_def conj_disj_distribs
disj_ac conj_ac conj_cancel_right disj_zero_left)

lemma xor_zero_right [simp]: "x ⊕ \<zero> = x"
by (simp only: xor_def compl_zero conj_one_right conj_zero_right disj_zero_right)

lemma xor_zero_left [simp]: "\<zero> ⊕ x = x"
by (subst xor_commute) (rule xor_zero_right)

lemma xor_one_right [simp]: "x ⊕ \<one> = ∼ x"
by (simp only: xor_def compl_one conj_zero_right conj_one_right disj_zero_left)

lemma xor_one_left [simp]: "\<one> ⊕ x = ∼ x"
by (subst xor_commute) (rule xor_one_right)

lemma xor_self [simp]: "x ⊕ x = \<zero>"
by (simp only: xor_def conj_cancel_right conj_cancel_left disj_zero_right)

lemma xor_left_self [simp]: "x ⊕ (x ⊕ y) = y"
by (simp only: xor_assoc [symmetric] xor_self xor_zero_left)

lemma xor_compl_left [simp]: "∼ x ⊕ y = ∼ (x ⊕ 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 [simp]: "x ⊕ ∼ y = ∼ (x ⊕ 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: "x ⊕ ∼ x = \<one>"
by (simp only: xor_compl_right xor_self compl_zero)

lemma xor_cancel_left: "∼ x ⊕ x = \<one>"
by (simp only: xor_compl_left xor_self compl_zero)

lemma conj_xor_distrib: "x \<sqinter> (y ⊕ z) = (x \<sqinter> y) ⊕ (x \<sqinter> z)"
proof -
have "(x \<sqinter> y \<sqinter> ∼ z) \<squnion> (x \<sqinter> ∼ y \<sqinter> z) =
(y \<sqinter> x \<sqinter> ∼ x) \<squnion> (z \<sqinter> x \<sqinter> ∼ x) \<squnion> (x \<sqinter> y \<sqinter> ∼ z) \<squnion> (x \<sqinter> ∼ y \<sqinter> z)"

by (simp only: conj_cancel_right conj_zero_right disj_zero_left)
thus "x \<sqinter> (y ⊕ z) = (x \<sqinter> y) ⊕ (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 ⊕ z) \<sqinter> x = (y \<sqinter> x) ⊕ (z \<sqinter> x)"
proof -
have "x \<sqinter> (y ⊕ z) = (x \<sqinter> y) ⊕ (x \<sqinter> z)"
by (rule conj_xor_distrib)
thus "(y ⊕ z) \<sqinter> x = (y \<sqinter> x) ⊕ (z \<sqinter> x)"
by (simp only: conj_commute)
qed

lemmas conj_xor_distribs =
conj_xor_distrib conj_xor_distrib2

end

end