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4.2 SOP ANDPOS FORMS                                                 133


                   SOP Term   Binary  Decimal  m i       SOP Term   Binary  Decimal  mi


                    AB C D    0000       0     rr> 0      AB C D    1000       8    m 8
                    A B C D   0001       1     m i        AB C D    1001       9    m g
                    A B C D   0010       2     m 2        AB C D    1010       10   m io
                    A I C D 0011         3     m 3        A B C D   1011       11   m n
                    A B C D   0100       4     m 4        AB C D    1100       12   m i2
                    A B C D   0101       5     m s        A B C D   1101       13   m  i3
                    A B C D   0110       6     m e        A B C D   1110       14   m  i4
                    A B C D   0111       7     m ?        A B C D   1111       15   m
                                                                                     !5
                  FIGURE 4.1
                 Minterm code table for four variables.


                 demonstrated by expanding Eq. (4.1) as follows:


                                 f(A, B, C)=AB + BC + ABC
                                          = AB(C + O + (A + A)BC + ABC
                                          = ABC + ABC + ABC + ABC + ABC
                                          = ni2 + m?, + ms + mj + m^
                                                     ,4,7).                          (4.3)


                 Note that the OR law X+X = 1 has been applied twice and that the two identical minterms
                 ABC are combined according to the OR law X + X = X.
                    The canonical truth table for Eqs. (4.3), shown in Fig. 4.2, is easily constructed from
                 the minterm code form. However, the truth table can also be constructed directly from the
                 original reduced form given by Eqs. (4.1). Notice that a logic 1 is placed in the / column
                 each time an AB = 01 occurs, each time a BC occurs, and for ABC. Thus, construction



                                               AB C   f

                                               00 0   0
                                               00 1    0
                                               01 0    1 m 2
                                               01 1    1 m 3
                                               1 0 0   1 m.
                                                           4
                                               10 1   0
                                               1 1 0  0
                                               1 1 1   1 m 7
                 FIGURE 4.2
                 Truth table for Eq. (4.3).
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