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the complement of individual variables  Application of DeMorgan’s theorem
              connected  by OR operator. For two       DeMorgan’s theorem  can be  used to
              variables, A and B and three variables,   simplify Boolean expression and create
              A, B and C, the AND product inverted     alternative  gates  from the  basic  gates.  Chapter One: Computer architecture
              theorem can be expressed as follows:     The following steps will guide you to
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               AB =  A B+   and  ABC =  A B C+  +  .   simplify a Boolean expression using
                                                       DeMorgan’s theorem  and Boolean
              Creating a truth table to prove          algebra laws.
              DeMorgan’s theorem
              The truth table is usually used to prove          Example 1.8
              different Boolean expressions. The truth   Simplify  the Boolean  expression
              table in this section is used to prove
              DeMorgan’s theorem.  The following         A BC+   using  DeMorgan’s theorem
              examples show the proof of DeMorgan’s      and Boolean algebra laws.
              theorem using the truth table.
                                                         Solution:
                      Example 1.7                        Step 1: Determine the terms you can

               Prove  DeMorgan’s theorem                         apply DeMorgan’s theorem
                A B+  =  . AB , by using truth table.            and treat each term as a single
                                                                 variable. Let  x =  A BC+   and
               Solution:                                         y =  BC
               Step 1:  The expression has two           Step 2:  Using DeMorgan’s theorem,
                        variables, A and B.                      the expression  y =  BC  will
               Step 2: Since the expression has two              be  y =  BC+  . Therefore,
                        variables, A and B; then, the
                        number of rows are  2 =  4               ABC+    =  AB C+  +
                                             2
               Step 3: Variables are A and B, and        Step 3:  Using DeMorgan’s theorem
                        every possible combination               the expression  A BC+  +    will
                        of 1 and 0 for each variable             be   =  A ..BC
                        is as in Table 1.10:
                                                         Step 4: Using double negation law,
               Table 1.10: Possible combinations for A           the expression  ..A BC   will be
                                   and B                         A ..BC

                 A   B   A   B A+B      A+ B   A.B

                 0   0   1    1    0     1      1              Example 1.9
                 0   1   1    0    1     1      1        Simplify  the Boolean  expression
                 1   0   0    1    1     1      1        ABC BC+        using    DeMorgan’s
                 1   1   0    0    1     0      0        theorem and Boolean algebra laws.


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               Student’s Book  Form Five



     Computer Science Form 5.indd   19                                                      23/07/2024   12:32
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