Question

Reduce the following Boolean expression using k-map.

A	B	C	F1
0	0	0	1
0	0	1	1
0	1	0	0
0	1	1	1
1	0	0	0
1	0	1	1
1	1	0	0
1	1	1	0

To SOP & POS form

Answer 1

For F1​(A, B, C)

From the truth table, F1 = 1 for:

000, 001, 011, 101

Therefore,

F1 ​= Σm(0, 1, 3, 5)

Minimized SOP form:

F1​ = A′B′ + A′C + B′C​

For POS, F1 = 0 for:

010, 100, 110, 111

F1 ​= ΠM(2, 4, 6, 7)

Minimized POS form:

F1 ​= (A′ + B′)(A′ + C)(B′ + C)​