Digital Electronics: Principles, Devices and Applications

Boolean Algebra and Simplification Techniques 199or by the sum of the remaining variables in the case of a product-of-sums expression will be redundant.The following example further illustrates the point:ABC + ACD + BCD + BCD + ACD = ABC + ACD + BCDIf we consider the first two terms of the Boolean expression, BCD becomes redundant. If we considerthe first and third terms of the given Boolean expression, ACD becomes redundant.Example 6.4Prove that ABCD + ABCD + ABCD + ABCD+ ABCDE + ABCDE + ABCDE canbe simplified to ABSolutionABCD + ABCD + ABCD + ABCD+ ABCDE + ABCDE + ABCDE= ABCD + ABCD + ABCD + ABCD= ABCD + CD + CD + CD = AB• ABCD appears in ABCDE, ABCD appears in ABCDE and ABCD appears inABCDE.• As a result, all three five-variable terms are redundant.• Also, variables C and D appear in all possible combinations and are therefore redundant.6.3.13 Theorem 13 (DeMorgan’s Theorem)(a) X 1 + X 2 + X 3 + + X n = X 1 X 2 X 3 X n (6.22)(b) X 1 X 2 X 3 X n = X 1 + X 2 + X 3 + + X n (6.23)According to the first theorem the complement of a sum equals the product of complements, whileaccording to the second theorem the complement of a product equals the sum of complements. Figures6.3(a) and (b) show logic diagram representations of De Morgan’s theorems. While the first theoremcan be interpreted to say that a multi-input NOR gate can be implemented as a multi-input bubbledAND gate, the second theorem, which is the dual of the first, can be interpreted to say that a multi-inputNAND gate can be implemented as a multi-input bubbled OR gate.DeMorgan’s theorem can be proved as follows. Let us assume that all variables are in a logic ‘0’state. In that caseLHS = X 1 + X 2 + X 3 + ··· +X n = 0 + 0 + 0 + ··· +0 = 0 = 1RHS = X 1 X 2 X 3 X n = 000 0 = 111 1 = 1Therefore, LHS = RHS.Now, let us assume that any one of the n variables, say X 1 , is in a logic HIGH state: