3860/CH3/EX3.23/Ex3_23.txt


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 F = m(0,3,4,5,6,7,8,10,11,14,15)

       A'B'  A'B  AB   AB'

 C'D'    1    1         1

 C'D          1    

 CD      1    1    1    1

 CD'          1    1    1

 The solution for F is

 F = A'B + CD + AC + B'C'D'