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Digital Circuits and Logic Design



                   Notes         Case 4.  For x = 1, y = 1,
                                                      x + xy  =  x
                                                    1 + 1 • 1  =  1

                                                       1 + 1  =  1
                                                          1  =  1
                                 Theorem, (14) can also be proved by factoring and using theorems (6) and (2) as follows:

                                                      x + xy  =  x (1 + y)
                                                             =  x 1   [using theorem (6)]
                                                             =  x    [using theorem (2)]
                                 All of these Boolean theorems can be useful in simplifying a logic expression—that is, in reducing
                                 the number of terms in the expression. When this is done, the reduced expression will produce a
                                 circuit that is less complex than the one which the original expression would have produced. For
                                 now, the following examples will serve to illustrate how the Boolean theorems can be applied.


                                             Simplify the expression y =  ABD +  ABD.

                                        Solution:
                                        Factor out the common variables  AB using theorem (13):
                                                             y  =  AB D( +  D)
                                                                                                  –
                                        Using theorem (8), the term in parentheses is equivalent to 1. Thus, y = AB • 1
                                                             y  =  AB   [using theorem (2)]


                                             Simplify z =  (A +  B )(A +  B ).

                                        Solution:
                                        The expression can be expanded by multiplying out the terms [theorem (13)]:
                                                             y  =  AB D( +  D)
                                                             z  =  AA•  +  AB +  BA + BB
                                                                                      •
                                                                               •
                                                                          •
                                        Invoking theorem (4), the term   AA•  = 0.  Also, B • B = B [theorem (3)]:
                                                                                 B
                                                             z  =  O +  A B +  B A + =  AB +  AB +  B
                                                                            •
                                                                       •
                                        Factoring out the variable B [theorem (13)], we have
                                                             z  =  BA(  +  A +  ) 1
                                        Finally, using theorems (2) and (6),
                                                             z  =  B


                                             Simplify x = ACD +   ABCD.

                                        Solution:
                                        Factoring out the common variables CD, we have
                                                             x  =  CD A(  +  AB)



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