src/HOL/Library/Boolean_Algebra.thy

bootstrap datatype_rep_proofs in Datatype.thy (avoids unchecked dynamic name references)

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

header {* Boolean Algebras *}

theory Boolean_Algebra
imports Main
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 [simp]: "\<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 [simp]: "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: "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 \<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