Boolean Algebra cont a ab a 1 ab Identity - PowerPoint PPT Presentation

Theorem: Absorption Law For every pair of elements a , b � B, �������������������� �� ����������� 1. a + a · b = a 2. a · ( a + b ) = a ��� � ������ Proof: (1) Boolean Algebra cont’ � � � � a ab a 1 ab Identity � � � � Distributivity a b 1 � � � � Commutativity a b 1 The digital abstraction � a � Theorem: For any a � B , 1 a + 1 = 1 � a Identity (2) duality. Proof: � � � � � � � � � � � � � � A a b c a b c (1) Let Theorem: Associative Law � � � � � � � � � � � � � � � � � A a b c a a b c b c Distributivity 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 � � � � � � � � � � � � � a b c a a a b c 2. a · ( b · c ) = ( 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

� � � � � � � � � � � � � � � � � A a b c a a b c b c � � � � � � � � � � � � � � c � � � � � � � � � a b c b c a b c b a b c � � � � � � � � � � � � � � c � � � � � � � � � a b c b c a b c b a b c c Same transitions Putting it all together: � � � � � � � � � � � � � � � � � � � � � � � Commutativity a b c b b a b c � � � � � � � A a b c a a b c b c � � � � � Distributivity b a b bc · before + � � � � � � � � � � � � � � � � � � � � � � � a b c a a b c b a b c c � � � Distributivity ba bb bc � � � a c b Idempotent Law ba b bc � ba � � � Absorption Law b � � � a b c � b � ba Commutativity � b Absorption Law Theorem 11: DeMorgan’s Law Also, � � � � � � � � � � � � � � � � � For every pair of elements a , b � B, A a b a b c c a b c 1. ( a + b ) ’ = a’ · b’ � � � � � � � � � � � � � � � � � � � � � � � a a b c b a b c c a b c 2. ( a · b ) ’ = a’ + b’ � � � � � a b c Proof: (1) We first prove that ( a+b ) is the complement of a’·b’ . � � � � � � � � � � A a b c a b c Thus, ( a+b ) ’ = a’·b’ By the definition of the complement and its uniqueness, it suffices to (2) Duality ( a+b )+( a’b’ ) = 1 and show: (i) ( a+b )( a’b’ ) = 0. (ii) (2) Duality ( a·b ) ’ = a’+b’

� ✁ ✁ � ✂ Commutativity � � � � � � � � � � � � � � � � � a b a b a b a b � � � � � � � � � � Distributivity � � � � � � � � � � � � Distributivity a b a b a b a a b b � � � � b � � � � � � a b a a b Commutativity Commutativity � � � � � � � � � � � � � � � � b a a a b b � � � � b � � � � � � b a a a b Associativity Associativity � � � � � � � � � � � � � � � � � � � � b a a a b b � � � � � � b a a a b b Commutativity a’ and b’ are the complements of � � � � � � � � � � � � � � � � � � a and b respectively b a b a a a b b 1 1 a’ and b’ are the complements of a and b respectively � � � � � � b a � 1 � 0 0 Theorem: For any a � B , 1 a + 1 = 1 Theorem: For any a � B , � 0 � 0 � 1 a · 0 = 0 Idempotent Law � 0 Idempotent Law Algebra of Sets Theorem: The algebra of sets is a Boolean algebra. Consider a set S. B = all the subsets of S (denoted by P(S) ) . Proof: “plus” � set-union By satisfying the axioms of Boolean algebra: “times” � set-intersection • B is a set of at least two elements � � � � � � � � M � � S � � , � P S P S , , For every non empty set S: |B| � 2 . Additive identity element – empty set Ø � � ) over B (functions ) . B B B • Closure of ( ) and ( Multiplicative identity element – the set S . � � X � X Y S P ( S , . ) by definition Complement of X � B: Y � P ( S ) by definition � � X S X \ � � � � X Y S X Y P S and ( ) by definition � � � � X Y S X Y P S and ( ) by definition

� � ✁ ✁ � A2. Distributivity of ( ) and ( ). A1. Cummutativity of ( ) and ( ). � � � � � � � � � � � � X Y Z X Y X Z � � � � � � X Y x : x X or x Y � � . � � � � � � x X Y Z � � � � Let Y X x x Y x X : or x � � � X x Y Z X � and An element lies in the union precisely when it lies in one of Y the two sets X and Y. Equally an element lies in the union Y � X precisely when it lies in one of the two sets X and Y. Hence, x � x � Y Z or � � � X Y Y X x � x � � � x � Y X Y x X Y If , We have and . Hence, � � � � � � X Y x x X x Y : and x � x � � � x � Z X Z x X Z If , We have and . Hence, � � � � � � Y X x x Y x X : and � � � � � � � � � � � � x X Y x X Z x X Y X Z or � � � X Y Y X � � � � � � � � � � � � X Y Z X Y X Z � This can be conducted in the same manner as . Taking (*) and (**) we get, We present an alternative way: � � � � � � � � � � � � X Y X Z X Y Z � � � � X Y X X Z X Definition of intersection and � � � � � � � � � � � � � � � � � X Y Z X Y X Z � � � � � X Y X Z X * � � X Y Y Also, definition of intersection � � � Distributivity of union over intersection can be conducted in the same manner. X Y Y Z � � Y Y Z definition of union � � � � � � � � � � � � X Y Z X Y X Z � � � X Z Y Z Similarly, � � � � � � � � � X Y X Z Y Z **