Explanation Given information: Let B be the Boolean algebra, with the operations, addition " + " and multiplication " . " . Concept used: a ¯ + a = a + a ¯ by the commutative law = 1 by the complement law for 1 And a ¯ ⋅ a = a ⋅ a ¯ by the commutative law = 0 by the complement law for 0 Calculation: Let B be the Boolean algebra, with the operations, addition " + " and multiplication " ⋅ " Suppose a and b are any elements of B . Then, a ¯ is the complement of a and b ¯ is the complement of b . The uniqueness of the complement law tells that, for all a and x in B if, a + x = 1 a . x = 0 ⇒ x = a ¯ To show that a . b ¯ = a ¯ + b ¯ , it is enough to show that ( a . b ) + ( a ¯ + b ¯ ) = 1 and, ( a . b ) . ( a ¯ + b ¯ ) = 0 Then use the uniqueness of complement law. Prove ( a . b ) . ( a ¯ + b ¯ ) = 1 ( a . b ) + ( a ¯ + b ¯ ) = ( a ¯ + b ¯ ) + ( a . b ) By the Commutative law for addition " + " = ( ( a ¯ + b ¯ ) + a ) . ( ( a ¯ + b ¯ ) + b ) By the Distributive law for addition " + " Over multiplication " . " = ( ( b ¯ + a ¯ ) + a ) . ( ( a ¯ + b ¯ ) + b ) By the Commutative law for addition " + " = ( b ¯ + ( a ¯ + a ) ) . ( a ¯ + ( b ¯ + b ) ) By the Associative law for addition " + " = ( b ¯ + ( a + a ¯ ) )

5th Edition

ISBN: 9780357097717

Author: EPP

Publisher: CENGAGE L

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Chapter 6.4, Problem 8ES

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To Prove:

For all a and b in B, a.b¯=a¯+b¯.

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Chapter 6.4, Problem 7ES

Chapter 6.4, Problem 9ES

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