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Given truth table
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A B C | f f'
0 0 0 | 1 1
0 0 1 | 0 0
0 1 0 | 1 0
0 1 1 | 1 0
1 0 0 | 1 0
1 0 1 | 0 0
1 1 0 | 0 1
1 1 1 | 0 1
f(A,B,C) = summation(1,2,3,4,5)
The complement of function is as given below
f'(A,B,C) = summation(0,6,7)
= A'B'C' + ABC' + ABC
= A'B'C' + AB
f = (A + B + C)(A' + B'
This is the reduced POS expression
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