src/HOL/Library/Boolean_Algebra.thy
 author nipkow Thu, 22 Oct 2020 08:39:08 +0200 changeset 72501 70b420065a07 parent 71042 400e9512f1d3 permissions -rw-r--r--
tuned names: t_ -> T_
```
(*  Title:      HOL/Library/Boolean_Algebra.thy
Author:     Brian Huffman
*)

section \<open>Boolean Algebras\<close>

theory Boolean_Algebra
imports Main
begin

locale boolean_algebra = conj: abel_semigroup "(\<^bold>\<sqinter>)" + disj: abel_semigroup "(\<^bold>\<squnion>)"
for conj :: "'a \<Rightarrow> 'a \<Rightarrow> 'a"  (infixr "\<^bold>\<sqinter>" 70)
and disj :: "'a \<Rightarrow> 'a \<Rightarrow> 'a"  (infixr "\<^bold>\<squnion>" 65) +
fixes compl :: "'a \<Rightarrow> 'a"  ("\<sim> _" [81] 80)
and zero :: "'a"  ("\<zero>")
and one  :: "'a"  ("\<one>")
assumes conj_disj_distrib: "x \<^bold>\<sqinter> (y \<^bold>\<squnion> z) = (x \<^bold>\<sqinter> y) \<^bold>\<squnion> (x \<^bold>\<sqinter> z)"
and disj_conj_distrib: "x \<^bold>\<squnion> (y \<^bold>\<sqinter> z) = (x \<^bold>\<squnion> y) \<^bold>\<sqinter> (x \<^bold>\<squnion> z)"
and conj_one_right: "x \<^bold>\<sqinter> \<one> = x"
and disj_zero_right: "x \<^bold>\<squnion> \<zero> = x"
and conj_cancel_right [simp]: "x \<^bold>\<sqinter> \<sim> x = \<zero>"
and disj_cancel_right [simp]: "x \<^bold>\<squnion> \<sim> x = \<one>"
begin

sublocale conj: semilattice_neutr "(\<^bold>\<sqinter>)" "\<one>"
proof
show "x \<^bold>\<sqinter> \<one> = x" for x
by (fact conj_one_right)
show "x \<^bold>\<sqinter> x = x" for x
proof -
have "x \<^bold>\<sqinter> x = (x \<^bold>\<sqinter> x) \<^bold>\<squnion> \<zero>"
also have "\<dots> = (x \<^bold>\<sqinter> x) \<^bold>\<squnion> (x \<^bold>\<sqinter> \<sim> x)"
by simp
also have "\<dots> = x \<^bold>\<sqinter> (x \<^bold>\<squnion> \<sim> x)"
by (simp only: conj_disj_distrib)
also have "\<dots> = x \<^bold>\<sqinter> \<one>"
by simp
also have "\<dots> = x"
finally show ?thesis .
qed
qed

sublocale disj: semilattice_neutr "(\<^bold>\<squnion>)" "\<zero>"
proof
show "x \<^bold>\<squnion> \<zero> = x" for x
by (fact disj_zero_right)
show "x \<^bold>\<squnion> x = x" for x
proof -
have "x \<^bold>\<squnion> x = (x \<^bold>\<squnion> x) \<^bold>\<sqinter> \<one>"
by simp
also have "\<dots> = (x \<^bold>\<squnion> x) \<^bold>\<sqinter> (x \<^bold>\<squnion> \<sim> x)"
by simp
also have "\<dots> = x \<^bold>\<squnion> (x \<^bold>\<sqinter> \<sim> x)"
by (simp only: disj_conj_distrib)
also have "\<dots> = x \<^bold>\<squnion> \<zero>"
by simp
also have "\<dots> = x"
finally show ?thesis .
qed
qed

subsection \<open>Complement\<close>

lemma complement_unique:
assumes 1: "a \<^bold>\<sqinter> x = \<zero>"
assumes 2: "a \<^bold>\<squnion> x = \<one>"
assumes 3: "a \<^bold>\<sqinter> y = \<zero>"
assumes 4: "a \<^bold>\<squnion> y = \<one>"
shows "x = y"
proof -
from 1 3 have "(a \<^bold>\<sqinter> x) \<^bold>\<squnion> (x \<^bold>\<sqinter> y) = (a \<^bold>\<sqinter> y) \<^bold>\<squnion> (x \<^bold>\<sqinter> y)"
by simp
then have "(x \<^bold>\<sqinter> a) \<^bold>\<squnion> (x \<^bold>\<sqinter> y) = (y \<^bold>\<sqinter> a) \<^bold>\<squnion> (y \<^bold>\<sqinter> x)"
then have "x \<^bold>\<sqinter> (a \<^bold>\<squnion> y) = y \<^bold>\<sqinter> (a \<^bold>\<squnion> x)"
with 2 4 have "x \<^bold>\<sqinter> \<one> = y \<^bold>\<sqinter> \<one>"
by simp
then show "x = y"
by simp
qed

lemma compl_unique: "x \<^bold>\<sqinter> y = \<zero> \<Longrightarrow> x \<^bold>\<squnion> y = \<one> \<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)
show "\<sim> x \<^bold>\<sqinter> x = \<zero>"
by (simp only: conj_cancel_right conj.commute)
show "\<sim> x \<^bold>\<squnion> x = \<one>"
by (simp only: disj_cancel_right disj.commute)
qed

lemma compl_eq_compl_iff [simp]: "\<sim> x = \<sim> y \<longleftrightarrow> x = y"
by (rule inj_eq [OF inj_on_inverseI]) (rule double_compl)

subsection \<open>Conjunction\<close>

lemma conj_zero_right [simp]: "x \<^bold>\<sqinter> \<zero> = \<zero>"
using conj.left_idem conj_cancel_right by fastforce

lemma compl_one [simp]: "\<sim> \<one> = \<zero>"
by (rule compl_unique [OF conj_zero_right disj_zero_right])

lemma conj_zero_left [simp]: "\<zero> \<^bold>\<sqinter> x = \<zero>"
by (subst conj.commute) (rule conj_zero_right)

lemma conj_cancel_left [simp]: "\<sim> x \<^bold>\<sqinter> x = \<zero>"
by (subst conj.commute) (rule conj_cancel_right)

lemma conj_disj_distrib2: "(y \<^bold>\<squnion> z) \<^bold>\<sqinter> x = (y \<^bold>\<sqinter> x) \<^bold>\<squnion> (z \<^bold>\<sqinter> x)"
by (simp only: conj.commute conj_disj_distrib)

lemmas conj_disj_distribs = conj_disj_distrib conj_disj_distrib2

lemma conj_assoc: "(x \<^bold>\<sqinter> y) \<^bold>\<sqinter> z = x \<^bold>\<sqinter> (y \<^bold>\<sqinter> z)"
by (fact ac_simps)

lemma conj_commute: "x \<^bold>\<sqinter> y = y \<^bold>\<sqinter> x"
by (fact ac_simps)

lemmas conj_left_commute = conj.left_commute
lemmas conj_ac = conj.assoc conj.commute conj.left_commute

lemma conj_one_left: "\<one> \<^bold>\<sqinter> x = x"
by (fact conj.left_neutral)

lemma conj_left_absorb: "x \<^bold>\<sqinter> (x \<^bold>\<sqinter> y) = x \<^bold>\<sqinter> y"
by (fact conj.left_idem)

lemma conj_absorb: "x \<^bold>\<sqinter> x = x"
by (fact conj.idem)

subsection \<open>Disjunction\<close>

interpretation dual: boolean_algebra "(\<^bold>\<squnion>)" "(\<^bold>\<sqinter>)" compl \<one> \<zero>
apply standard
apply (rule disj_conj_distrib)
apply (rule conj_disj_distrib)
apply simp_all
done

lemma compl_zero [simp]: "\<sim> \<zero> = \<one>"
by (fact dual.compl_one)

lemma disj_one_right [simp]: "x \<^bold>\<squnion> \<one> = \<one>"
by (fact dual.conj_zero_right)

lemma disj_one_left [simp]: "\<one> \<^bold>\<squnion> x = \<one>"
by (fact dual.conj_zero_left)

lemma disj_cancel_left [simp]: "\<sim> x \<^bold>\<squnion> x = \<one>"
by (rule dual.conj_cancel_left)

lemma disj_conj_distrib2: "(y \<^bold>\<sqinter> z) \<^bold>\<squnion> x = (y \<^bold>\<squnion> x) \<^bold>\<sqinter> (z \<^bold>\<squnion> x)"
by (rule dual.conj_disj_distrib2)

lemmas disj_conj_distribs = disj_conj_distrib disj_conj_distrib2

lemma disj_assoc: "(x \<^bold>\<squnion> y) \<^bold>\<squnion> z = x \<^bold>\<squnion> (y \<^bold>\<squnion> z)"
by (fact ac_simps)

lemma disj_commute: "x \<^bold>\<squnion> y = y \<^bold>\<squnion> x"
by (fact ac_simps)

lemmas disj_left_commute = disj.left_commute

lemmas disj_ac = disj.assoc disj.commute disj.left_commute

lemma disj_zero_left: "\<zero> \<^bold>\<squnion> x = x"
by (fact disj.left_neutral)

lemma disj_left_absorb: "x \<^bold>\<squnion> (x \<^bold>\<squnion> y) = x \<^bold>\<squnion> y"
by (fact disj.left_idem)

lemma disj_absorb: "x \<^bold>\<squnion> x = x"
by (fact disj.idem)

subsection \<open>De Morgan's Laws\<close>

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

lemma de_Morgan_disj [simp]: "\<sim> (x \<^bold>\<squnion> y) = \<sim> x \<^bold>\<sqinter> \<sim> y"
using dual.boolean_algebra_axioms by (rule boolean_algebra.de_Morgan_conj)

subsection \<open>Symmetric Difference\<close>

definition xor :: "'a \<Rightarrow> 'a \<Rightarrow> 'a"  (infixr "\<oplus>" 65)
where "x \<oplus> y = (x \<^bold>\<sqinter> \<sim> y) \<^bold>\<squnion> (\<sim> x \<^bold>\<sqinter> y)"

sublocale xor: comm_monoid xor \<zero>
proof
fix x y z :: 'a
let ?t = "(x \<^bold>\<sqinter> y \<^bold>\<sqinter> z) \<^bold>\<squnion> (x \<^bold>\<sqinter> \<sim> y \<^bold>\<sqinter> \<sim> z) \<^bold>\<squnion> (\<sim> x \<^bold>\<sqinter> y \<^bold>\<sqinter> \<sim> z) \<^bold>\<squnion> (\<sim> x \<^bold>\<sqinter> \<sim> y \<^bold>\<sqinter> z)"
have "?t \<^bold>\<squnion> (z \<^bold>\<sqinter> x \<^bold>\<sqinter> \<sim> x) \<^bold>\<squnion> (z \<^bold>\<sqinter> y \<^bold>\<sqinter> \<sim> y) = ?t \<^bold>\<squnion> (x \<^bold>\<sqinter> y \<^bold>\<sqinter> \<sim> y) \<^bold>\<squnion> (x \<^bold>\<sqinter> z \<^bold>\<sqinter> \<sim> z)"
by (simp only: conj_cancel_right conj_zero_right)
then show "(x \<oplus> y) \<oplus> z = x \<oplus> (y \<oplus> z)"
by (simp only: xor_def de_Morgan_disj de_Morgan_conj double_compl)
(simp only: conj_disj_distribs conj_ac disj_ac)
show "x \<oplus> y = y \<oplus> x"
by (simp only: xor_def conj_commute disj_commute)
show "x \<oplus> \<zero> = x"
qed

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 \<oplus> y = (x \<^bold>\<squnion> y) \<^bold>\<sqinter> (\<sim> x \<^bold>\<squnion> \<sim> y)"
using conj.commute conj_disj_distrib2 disj.commute xor_def by auto

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 [simp]: "\<sim> x \<oplus> y = \<sim> (x \<oplus> y)"
by (metis xor_assoc xor_one_left)

lemma xor_compl_right [simp]: "x \<oplus> \<sim> y = \<sim> (x \<oplus> y)"
using xor_commute xor_compl_left by auto

lemma xor_cancel_right: "x \<oplus> \<sim> x = \<one>"
by (simp only: xor_compl_right xor_self compl_zero)

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

lemma conj_xor_distrib: "x \<^bold>\<sqinter> (y \<oplus> z) = (x \<^bold>\<sqinter> y) \<oplus> (x \<^bold>\<sqinter> z)"
proof -
have *: "(x \<^bold>\<sqinter> y \<^bold>\<sqinter> \<sim> z) \<^bold>\<squnion> (x \<^bold>\<sqinter> \<sim> y \<^bold>\<sqinter> z) =
(y \<^bold>\<sqinter> x \<^bold>\<sqinter> \<sim> x) \<^bold>\<squnion> (z \<^bold>\<sqinter> x \<^bold>\<sqinter> \<sim> x) \<^bold>\<squnion> (x \<^bold>\<sqinter> y \<^bold>\<sqinter> \<sim> z) \<^bold>\<squnion> (x \<^bold>\<sqinter> \<sim> y \<^bold>\<sqinter> z)"
by (simp only: conj_cancel_right conj_zero_right disj_zero_left)
then show "x \<^bold>\<sqinter> (y \<oplus> z) = (x \<^bold>\<sqinter> y) \<oplus> (x \<^bold>\<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) \<^bold>\<sqinter> x = (y \<^bold>\<sqinter> x) \<oplus> (z \<^bold>\<sqinter> x)"