--- a/src/HOL/Library/Boolean_Algebra.thy Sat Apr 20 18:02:22 2019 +0000
+++ b/src/HOL/Library/Boolean_Algebra.thy Sat Apr 20 18:02:22 2019 +0000
@@ -8,33 +8,33 @@
imports Main
begin
-locale boolean_algebra = conj: abel_semigroup "(\<sqinter>)" + disj: abel_semigroup "(\<squnion>)"
- for conj :: "'a \<Rightarrow> 'a \<Rightarrow> 'a" (infixr "\<sqinter>" 70)
- and disj :: "'a \<Rightarrow> 'a \<Rightarrow> 'a" (infixr "\<squnion>" 65) +
+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 \<sqinter> (y \<squnion> z) = (x \<sqinter> y) \<squnion> (x \<sqinter> z)"
- and disj_conj_distrib: "x \<squnion> (y \<sqinter> z) = (x \<squnion> y) \<sqinter> (x \<squnion> z)"
- and conj_one_right: "x \<sqinter> \<one> = x"
- and disj_zero_right: "x \<squnion> \<zero> = x"
- and conj_cancel_right [simp]: "x \<sqinter> \<sim> x = \<zero>"
- and disj_cancel_right [simp]: "x \<squnion> \<sim> x = \<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 "(\<sqinter>)" "\<one>"
+sublocale conj: semilattice_neutr "(\<^bold>\<sqinter>)" "\<one>"
proof
- show "x \<sqinter> \<one> = x" for x
+ show "x \<^bold>\<sqinter> \<one> = x" for x
by (fact conj_one_right)
- show "x \<sqinter> x = x" for x
+ show "x \<^bold>\<sqinter> x = x" for x
proof -
- have "x \<sqinter> x = (x \<sqinter> x) \<squnion> \<zero>"
+ have "x \<^bold>\<sqinter> x = (x \<^bold>\<sqinter> x) \<^bold>\<squnion> \<zero>"
by (simp add: disj_zero_right)
- also have "\<dots> = (x \<sqinter> x) \<squnion> (x \<sqinter> \<sim> x)"
+ also have "\<dots> = (x \<^bold>\<sqinter> x) \<^bold>\<squnion> (x \<^bold>\<sqinter> \<sim> x)"
by simp
- also have "\<dots> = x \<sqinter> (x \<squnion> \<sim> x)"
+ also have "\<dots> = x \<^bold>\<sqinter> (x \<^bold>\<squnion> \<sim> x)"
by (simp only: conj_disj_distrib)
- also have "\<dots> = x \<sqinter> \<one>"
+ also have "\<dots> = x \<^bold>\<sqinter> \<one>"
by simp
also have "\<dots> = x"
by (simp add: conj_one_right)
@@ -42,19 +42,19 @@
qed
qed
-sublocale disj: semilattice_neutr "(\<squnion>)" "\<zero>"
+sublocale disj: semilattice_neutr "(\<^bold>\<squnion>)" "\<zero>"
proof
- show "x \<squnion> \<zero> = x" for x
+ show "x \<^bold>\<squnion> \<zero> = x" for x
by (fact disj_zero_right)
- show "x \<squnion> x = x" for x
+ show "x \<^bold>\<squnion> x = x" for x
proof -
- have "x \<squnion> x = (x \<squnion> x) \<sqinter> \<one>"
+ have "x \<^bold>\<squnion> x = (x \<^bold>\<squnion> x) \<^bold>\<sqinter> \<one>"
by simp
- also have "\<dots> = (x \<squnion> x) \<sqinter> (x \<squnion> \<sim> x)"
+ also have "\<dots> = (x \<^bold>\<squnion> x) \<^bold>\<sqinter> (x \<^bold>\<squnion> \<sim> x)"
by simp
- also have "\<dots> = x \<squnion> (x \<sqinter> \<sim> x)"
+ also have "\<dots> = x \<^bold>\<squnion> (x \<^bold>\<sqinter> \<sim> x)"
by (simp only: disj_conj_distrib)
- also have "\<dots> = x \<squnion> \<zero>"
+ also have "\<dots> = x \<^bold>\<squnion> \<zero>"
by simp
also have "\<dots> = x"
by (simp add: disj_zero_right)
@@ -66,32 +66,32 @@
subsection \<open>Complement\<close>
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>"
+ 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 \<sqinter> x) \<squnion> (x \<sqinter> y) = (a \<sqinter> y) \<squnion> (x \<sqinter> y)"
+ 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 \<sqinter> a) \<squnion> (x \<sqinter> y) = (y \<sqinter> a) \<squnion> (y \<sqinter> x)"
+ then have "(x \<^bold>\<sqinter> a) \<^bold>\<squnion> (x \<^bold>\<sqinter> y) = (y \<^bold>\<sqinter> a) \<^bold>\<squnion> (y \<^bold>\<sqinter> x)"
by (simp add: ac_simps)
- then have "x \<sqinter> (a \<squnion> y) = y \<sqinter> (a \<squnion> x)"
+ then have "x \<^bold>\<sqinter> (a \<^bold>\<squnion> y) = y \<^bold>\<sqinter> (a \<^bold>\<squnion> x)"
by (simp add: conj_disj_distrib)
- with 2 4 have "x \<sqinter> \<one> = y \<sqinter> \<one>"
+ 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 \<sqinter> y = \<zero> \<Longrightarrow> x \<squnion> y = \<one> \<Longrightarrow> \<sim> x = y"
+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 \<sqinter> x = \<zero>"
+ show "\<sim> x \<^bold>\<sqinter> x = \<zero>"
by (simp only: conj_cancel_right conj.commute)
- show "\<sim> x \<squnion> x = \<one>"
+ show "\<sim> x \<^bold>\<squnion> x = \<one>"
by (simp only: disj_cancel_right disj.commute)
qed
@@ -101,13 +101,13 @@
subsection \<open>Conjunction\<close>
-lemma conj_zero_right [simp]: "x \<sqinter> \<zero> = \<zero>"
+lemma conj_zero_right [simp]: "x \<^bold>\<sqinter> \<zero> = \<zero>"
proof -
- from conj_cancel_right have "x \<sqinter> \<zero> = x \<sqinter> (x \<sqinter> \<sim> x)"
+ from conj_cancel_right have "x \<^bold>\<sqinter> \<zero> = x \<^bold>\<sqinter> (x \<^bold>\<sqinter> \<sim> x)"
by simp
- also from conj_assoc have "\<dots> = (x \<sqinter> x) \<sqinter> \<sim> x"
+ also from conj_assoc have "\<dots> = (x \<^bold>\<sqinter> x) \<^bold>\<sqinter> \<sim> x"
by (simp only: ac_simps)
- also from conj_absorb have "\<dots> = x \<sqinter> \<sim> x"
+ also from conj_absorb have "\<dots> = x \<^bold>\<sqinter> \<sim> x"
by simp
also have "\<dots> = \<zero>"
by simp
@@ -117,39 +117,39 @@
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>"
+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 \<sqinter> x = \<zero>"
+lemma conj_cancel_left [simp]: "\<sim> x \<^bold>\<sqinter> x = \<zero>"
by (subst conj.commute) (rule conj_cancel_right)
-lemma conj_disj_distrib2: "(y \<squnion> z) \<sqinter> x = (y \<sqinter> x) \<squnion> (z \<sqinter> x)"
+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 \<sqinter> y) \<sqinter> z = x \<sqinter> (y \<sqinter> z)"
+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 \<sqinter> y = y \<sqinter> x"
+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> \<sqinter> x = x"
+lemma conj_one_left: "\<one> \<^bold>\<sqinter> x = x"
by (fact conj.left_neutral)
-lemma conj_left_absorb: "x \<sqinter> (x \<sqinter> y) = x \<sqinter> y"
+lemma conj_left_absorb: "x \<^bold>\<sqinter> (x \<^bold>\<sqinter> y) = x \<^bold>\<sqinter> y"
by (fact conj.left_idem)
-lemma conj_absorb: "x \<sqinter> x = x"
+lemma conj_absorb: "x \<^bold>\<sqinter> x = x"
by (fact conj.idem)
subsection \<open>Disjunction\<close>
-interpretation dual: boolean_algebra "(\<squnion>)" "(\<sqinter>)" compl \<one> \<zero>
+interpretation dual: boolean_algebra "(\<^bold>\<squnion>)" "(\<^bold>\<sqinter>)" compl \<one> \<zero>
apply standard
apply (rule disj_conj_distrib)
apply (rule conj_disj_distrib)
@@ -159,73 +159,73 @@
lemma compl_zero [simp]: "\<sim> \<zero> = \<one>"
by (fact dual.compl_one)
-lemma disj_one_right [simp]: "x \<squnion> \<one> = \<one>"
+lemma disj_one_right [simp]: "x \<^bold>\<squnion> \<one> = \<one>"
by (fact dual.conj_zero_right)
-lemma disj_one_left [simp]: "\<one> \<squnion> x = \<one>"
+lemma disj_one_left [simp]: "\<one> \<^bold>\<squnion> x = \<one>"
by (fact dual.conj_zero_left)
-lemma disj_cancel_left [simp]: "\<sim> x \<squnion> x = \<one>"
+lemma disj_cancel_left [simp]: "\<sim> x \<^bold>\<squnion> x = \<one>"
by (rule dual.conj_cancel_left)
-lemma disj_conj_distrib2: "(y \<sqinter> z) \<squnion> x = (y \<squnion> x) \<sqinter> (z \<squnion> x)"
+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 \<squnion> y) \<squnion> z = x \<squnion> (y \<squnion> z)"
+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 \<squnion> y = y \<squnion> x"
+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> \<squnion> x = x"
+lemma disj_zero_left: "\<zero> \<^bold>\<squnion> x = x"
by (fact disj.left_neutral)
-lemma disj_left_absorb: "x \<squnion> (x \<squnion> y) = x \<squnion> y"
+lemma disj_left_absorb: "x \<^bold>\<squnion> (x \<^bold>\<squnion> y) = x \<^bold>\<squnion> y"
by (fact disj.left_idem)
-lemma disj_absorb: "x \<squnion> x = x"
+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 \<sqinter> y) = \<sim> x \<squnion> \<sim> y"
+lemma de_Morgan_conj [simp]: "\<sim> (x \<^bold>\<sqinter> y) = \<sim> x \<^bold>\<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)"
+ 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 \<sqinter> (x \<sqinter> \<sim> x)) \<squnion> (x \<sqinter> (y \<sqinter> \<sim> y))"
+ 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 \<sqinter> y) \<sqinter> (\<sim> x \<squnion> \<sim> y) = \<zero>"
+ 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 \<sqinter> y) \<squnion> (\<sim> x \<squnion> \<sim> y) = (x \<squnion> (\<sim> x \<squnion> \<sim> y)) \<sqinter> (y \<squnion> (\<sim> x \<squnion> \<sim> y))"
+ 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 \<squnion> (x \<squnion> \<sim> x)) \<sqinter> (\<sim> x \<squnion> (y \<squnion> \<sim> y))"
+ 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 \<sqinter> y) \<squnion> (\<sim> x \<squnion> \<sim> y) = \<one>"
+ 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 \<squnion> y) = \<sim> x \<sqinter> \<sim> y"
+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 \<sqinter> \<sim> y) \<squnion> (\<sim> x \<sqinter> y)"
+ where "x \<oplus> y = (x \<^bold>\<sqinter> \<sim> y) \<^bold>\<squnion> (\<sim> x \<^bold>\<sqinter> y)"
sublocale xor: abel_semigroup xor
proof
fix x y z :: 'a
- 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)"
+ 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)
@@ -240,7 +240,7 @@
lemmas xor_ac = xor.assoc xor.commute xor.left_commute
-lemma xor_def2: "x \<oplus> y = (x \<squnion> y) \<sqinter> (\<sim> x \<squnion> \<sim> y)"
+lemma xor_def2: "x \<oplus> y = (x \<^bold>\<squnion> y) \<^bold>\<sqinter> (\<sim> x \<^bold>\<squnion> \<sim> 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 \<oplus> \<zero> = x"
@@ -283,22 +283,22 @@
lemma xor_cancel_left: "\<sim> x \<oplus> x = \<one>"
by (simp only: xor_compl_left xor_self compl_zero)
-lemma conj_xor_distrib: "x \<sqinter> (y \<oplus> z) = (x \<sqinter> y) \<oplus> (x \<sqinter> z)"
+lemma conj_xor_distrib: "x \<^bold>\<sqinter> (y \<oplus> z) = (x \<^bold>\<sqinter> y) \<oplus> (x \<^bold>\<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)"
+ 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 \<sqinter> (y \<oplus> z) = (x \<sqinter> y) \<oplus> (x \<sqinter> z)"
+ 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) \<sqinter> x = (y \<sqinter> x) \<oplus> (z \<sqinter> x)"
+lemma conj_xor_distrib2: "(y \<oplus> z) \<^bold>\<sqinter> x = (y \<^bold>\<sqinter> x) \<oplus> (z \<^bold>\<sqinter> x)"
proof -
- have "x \<sqinter> (y \<oplus> z) = (x \<sqinter> y) \<oplus> (x \<sqinter> z)"
+ have "x \<^bold>\<sqinter> (y \<oplus> z) = (x \<^bold>\<sqinter> y) \<oplus> (x \<^bold>\<sqinter> z)"
by (rule conj_xor_distrib)
- then show "(y \<oplus> z) \<sqinter> x = (y \<sqinter> x) \<oplus> (z \<sqinter> x)"
+ then show "(y \<oplus> z) \<^bold>\<sqinter> x = (y \<^bold>\<sqinter> x) \<oplus> (z \<^bold>\<sqinter> x)"
by (simp only: conj_commute)
qed