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theory Boolean_Algebra(* ID: $Id: Boolean_Algebra.thy,v 1.6 2007/12/18 13:37:01 haftmann Exp $ Author: Brian Huffman Boolean algebras as locales. *) header {* Boolean Algebras *} theory Boolean_Algebra imports ATP_Linkup 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>" 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: "[|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)" begin 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_commute: "x ⊕ y = y ⊕ x" by (simp only: xor_def conj_commute disj_commute) lemma xor_assoc: "(x ⊕ y) ⊕ z = x ⊕ (y ⊕ z)" proof - 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 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 ⊕ \<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: "∼ 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: "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 [simp]: "x ⊕ ∼ x = \<one>" by (simp only: xor_compl_right xor_self compl_zero) lemma xor_cancel_left [simp]: "∼ x ⊕ x = \<one>" by (subst xor_commute) (rule xor_cancel_right) 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
lemma disj_ac:
(x \<squnion> y) \<squnion> z = x \<squnion> y \<squnion> z
x \<squnion> y = y \<squnion> x
x \<squnion> y \<squnion> z = y \<squnion> x \<squnion> z
lemma conj_ac:
(x \<sqinter> y) \<sqinter> z = x \<sqinter> y \<sqinter> z
x \<sqinter> y = y \<sqinter> x
x \<sqinter> y \<sqinter> z = y \<sqinter> x \<sqinter> z
lemma dual:
boolean op \<squnion> op \<sqinter> compl \<one> \<zero>
lemma complement_unique:
[| a \<sqinter> x = \<zero>; a \<squnion> x = \<one>; a \<sqinter> y = \<zero>;
a \<squnion> y = \<one> |]
==> x = y
lemma compl_unique:
[| x \<sqinter> y = \<zero>; x \<squnion> y = \<one> |] ==> ∼ x = y
lemma double_compl:
∼ (∼ x) = x
lemma compl_eq_compl_iff:
(∼ x = ∼ y) = (x = y)
lemma conj_absorb:
x \<sqinter> x = x
lemma conj_zero_right:
x \<sqinter> \<zero> = \<zero>
lemma compl_one:
∼ \<one> = \<zero>
lemma conj_zero_left:
\<zero> \<sqinter> x = \<zero>
lemma conj_one_left:
\<one> \<sqinter> x = x
lemma conj_cancel_left:
∼ x \<sqinter> x = \<zero>
lemma conj_left_absorb:
x \<sqinter> x \<sqinter> y = x \<sqinter> y
lemma conj_disj_distrib2:
(y \<squnion> z) \<sqinter> x = y \<sqinter> x \<squnion> z \<sqinter> x
lemma conj_disj_distribs:
x \<sqinter> (y \<squnion> z) = x \<sqinter> y \<squnion> x \<sqinter> z
(y \<squnion> z) \<sqinter> x = y \<sqinter> x \<squnion> z \<sqinter> x
lemma disj_absorb:
x \<squnion> x = x
lemma disj_one_right:
x \<squnion> \<one> = \<one>
lemma compl_zero:
∼ \<zero> = \<one>
lemma disj_zero_left:
\<zero> \<squnion> x = x
lemma disj_one_left:
\<one> \<squnion> x = \<one>
lemma disj_cancel_left:
∼ x \<squnion> x = \<one>
lemma disj_left_absorb:
x \<squnion> x \<squnion> y = x \<squnion> y
lemma disj_conj_distrib2:
y \<sqinter> z \<squnion> x = (y \<squnion> x) \<sqinter> (z \<squnion> x)
lemma disj_conj_distribs:
x \<squnion> y \<sqinter> z = (x \<squnion> y) \<sqinter> (x \<squnion> z)
y \<sqinter> z \<squnion> x = (y \<squnion> x) \<sqinter> (z \<squnion> x)
lemma de_Morgan_conj:
∼ (x \<sqinter> y) = ∼ x \<squnion> ∼ y
lemma de_Morgan_disj:
∼ (x \<squnion> y) = ∼ x \<sqinter> ∼ y
lemma xor_def2:
x ⊕ y = (x \<squnion> y) \<sqinter> (∼ x \<squnion> ∼ y)
lemma xor_commute:
x ⊕ y = y ⊕ x
lemma xor_assoc:
(x ⊕ y) ⊕ z = x ⊕ y ⊕ z
lemma xor_ac:
(x ⊕ y) ⊕ z = x ⊕ y ⊕ z
x ⊕ y = y ⊕ x
x ⊕ y ⊕ z = y ⊕ x ⊕ z
lemma xor_zero_right:
x ⊕ \<zero> = x
lemma xor_zero_left:
\<zero> ⊕ x = x
lemma xor_one_right:
x ⊕ \<one> = ∼ x
lemma xor_one_left:
\<one> ⊕ x = ∼ x
lemma xor_self:
x ⊕ x = \<zero>
lemma xor_left_self:
x ⊕ x ⊕ y = y
lemma xor_compl_left:
∼ x ⊕ y = ∼ (x ⊕ y)
lemma xor_compl_right:
x ⊕ ∼ y = ∼ (x ⊕ y)
lemma xor_cancel_right:
x ⊕ ∼ x = \<one>
lemma xor_cancel_left:
∼ x ⊕ x = \<one>
lemma conj_xor_distrib:
x \<sqinter> (y ⊕ z) = x \<sqinter> y ⊕ x \<sqinter> z
lemma conj_xor_distrib2:
(y ⊕ z) \<sqinter> x = y \<sqinter> x ⊕ z \<sqinter> x
lemma conj_xor_distribs:
x \<sqinter> (y ⊕ z) = x \<sqinter> y ⊕ x \<sqinter> z
(y ⊕ z) \<sqinter> x = y \<sqinter> x ⊕ z \<sqinter> x