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Boolean Algebra cont The digital abstraction Theorem: Absorption - PowerPoint PPT Presentation

Boolean Algebra cont The digital abstraction Theorem: Absorption Law For every pair of elements a , b


  1. �������������������� �� ����������� ��� � ������ Boolean Algebra cont’ The digital abstraction

  2. Theorem: Absorption Law For every pair of elements a , b � B, 1. a + a · b = a 2. a · ( a + b ) = a Proof: (1) � � � � a ab a ab 1 Identity � � � � Distributivity a b 1 � � � � Commutativity a b 1 � a � Theorem: For any a � B , 1 a + 1 = 1 � a Identity (2) duality.

  3. Theorem: Associative Law In a Boolean algebra, each of the binary operations ( + ) and ( · ) is associative. That is, for every a , b , c � B, 1. a + ( b + c ) = ( a + b ) + c 2. a · ( b · c ) = ( a · b ) · c

  4. Proof: � � � � � � � � � � � � � � A a b c a b c (1) Let � � � � � � � � � � � � � � � � � A a b c a a b c b c Distributivity � � � � � � � � � � � � � a b c a a a b c Commutativity � � � � � a a b ac Distributivity � � � aa ab ac Distributivity � � � a ab ac Idempotent Law � a � ac Absorption Law � a Absorption Law

  5. � � � � � � � � � � � � � � � � � A a b c a a b c b c � � � � � � � c � � � � � � � � � � � � � � � � a b c b c a b c b a b c � � � � � � � � � � � � � Commutativity a b c b b a b c � � � � � Distributivity b a b bc � � � Distributivity ba bb bc � � � Idempotent Law ba b bc � ba � b Absorption Law � b � ba Commutativity � b Absorption Law

  6. � � � � � � � c � � � � � � � � � � � � � � � � a b c b c a b c b a b c c Same transitions Putting it all together: � � � � � � � � � � � � � � � � � A a b c a a b c b c · before + � � � � � � � � � � � � � � � � � � � � � � � a b c a a b c b a b c c a c b � � � � � a b c

  7. Also, � � � � � � � � � � � � � � � � � A a b a b c c a b c � � � � � � � � � � � � � � � � � � � � � � � a a b c b a b c c a b c � � � � � a b c � � � � � � � � � � A a b c a b c (2) Duality

  8. Theorem 11: DeMorgan’s Law For every pair of elements a , b � B, 1. ( a + b ) ’ = a’ · b’ 2. ( a · b ) ’ = a’ + b’ Proof: (1) We first prove that ( a+b ) is the complement of a’·b’ . Thus, ( a+b ) ’ = a’·b’ By the definition of the complement and its uniqueness, it suffices to ( a+b )+( a’b’ ) = 1 and show: (i) ( a+b )( a’b’ ) = 0. (ii) ( a·b ) ’ = a’+b’ (2) Duality

  9. � � � � � � � � � � Distributivity � � � � � � � � � � � � a b a b a b a a b b Commutativity � � � � � � � � � � � � � � � � b a a a b b Associativity � � � � � � � � � � � � � � � � b a a a b b a’ and b’ are the complements of � � � � � � � � b a a and b respectively 1 1 � 1 � Theorem: For any a � B , 1 a + 1 = 1 � 1 Idempotent Law

  10. Commutativity � � � � � � � � � � � � � � � � � a b a b a b a b Distributivity � � � � b � � � � � � a b a a b Commutativity � � � � b � � � � � � b a a a b Associativity � � � � � � � � � � b a a a b b Commutativity � � � � � � � � � � b a a a b b a’ and b’ are the complements of a and b respectively � � � � � � b a 0 0 Theorem: For any a � B , � 0 � 0 a · 0 = 0 � 0 Idempotent Law

  11. � Algebra of Sets Consider a set S. B = all the subsets of S (denoted by P(S) ) . “plus” � set-union “times” � set-intersection � � � � M � � , � P S , Additive identity element – empty set Ø Multiplicative identity element – the set S . Complement of X � B: � � X S X \

  12. � Theorem: The algebra of sets is a Boolean algebra. Proof: By satisfying the axioms of Boolean algebra: • B is a set of at least two elements � � � � � S � P S , For every non empty set S: |B| � 2 . � � ) and ( ) over B (functions ) . B B B • Closure of ( � � X � X Y S , . P ( S ) by definition Y � P ( S ) by definition � � � � X Y S X Y P S and ( ) by definition � � � � X Y S X Y P S and ( ) by definition

  13. � A1. Cummutativity of ( ) and ( ). � � � � � � X Y x x X x Y : or � � � � � � Y X x x Y x X : or X � An element lies in the union precisely when it lies in one of Y Y � the two sets X and Y. Equally an element lies in the union X precisely when it lies in one of the two sets X and Y. Hence, � � � X Y Y X � � � � � � X Y x x X x Y : and � � � � � � Y X x x Y x X : and � � � X Y Y X

  14. � A2. Distributivity of ( ) and ( ). � � � � � � � � � � � � X Y Z X Y X Z � � . � � � � x X Y Z Let x � � � X x Y Z and x � x � Y Z or x � x � � � x � X Y x X Y Y If , We have and . Hence, x � x � � � x � X Z x X Z Z If , We have and . Hence, � � � � � � � � � � � � x X Y X Z x X Y x X Z or � � � � � � � � � � � � X Y Z X Y X Z

  15. � � This can be conducted in the same manner as . We present an alternative way: � � � � X Y X X Z X Definition of intersection and � � � � � � � � X Y X Z X * � � X Y Y Also, definition of intersection � � � X Y Y Z � � Y Y Z definition of union � � � X Z Y Z Similarly, � � � � � � � � � X Y X Z Y Z **

  16. Taking (*) and (**) we get, � � � � � � � � � � � � X Y X Z X Y Z � � � � � � � � � � � � � X Y Z X Y X Z � Distributivity of union over intersection can be conducted in the same manner. � � � � � � � � � � � � X Y Z X Y X Z

  17. A3. Existence of additive and multiplicative identity element. � � � � � � � � � X S X X X . - additive identity � � � � � � X S X S S X X S . - multiplica tive identity A4. Existence of the complement. � � � � X B X S X . \ � � � � � � � � � � X B X S X S X X S . \ \ � � � � � � � � � � � X B X S X S X X . \ \ All axioms are satisfied Algebra of sets is Boolean algebra.

  18. Boolean expression - Recursive definition: base: 0 , 1 , a � B – expressions. recursion step: Let E 1 and E 2 be Boolean expressions. Then, E 1 ’ ( E 1 + E 2 ) ( E 1 · E 2 ) Dual transformation - Recursive definition: Dual: expressions expressions base: 0 1 1 0 a , a � B \{0,1} a recursion step: Let E 1 and E 2 be Boolean expressions. Then, E 1 ’ [dual(E 1 )]’ ( E 1 + E 2 ) [ dual(E 1 ) · dual(E 2 ) ] ( E 1 · E 2 ) [ dual(E 1 ) + dual(E 2 ) ]

  19. Let f d be the dual of a function f ( x 1 , x 2 , … , x n ) f d = f’ ( x 1 ’ , x 2 ’ , … , x n ’ ) Lemma: In switching algebra, Proof: Let f ( x 1 , x 2 , … , x n ) be a Boolean expression. We show that applying the complement on the whole expression together with replacing each variable by it’s complement, yields the dual transformation definition. Induction basis: 0 , 1 – expressions. � f � � � � � � � � � � � � � � � � � � � � � � � � � f x f x f x x x 0 , 2 , , 0 1 n 1 d f � � � � � � � � � � � � � � � � � � � � � � � � � � f x f x f x x x 1 , 2 , , 1 0 n 1 d

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