Given $f_1$, $f_3$ and $f$ in canonical sum of products form (in decimal) for the given circuit find $f_2$.

Given $f_1$, $f_3$ and $f$ in canonical sum of products form (in decimal) for the circuit 

$f_1 = \sum m(4, 5, 6, 7, 8)$, 

$f_3 = \sum m(1, 6, 15)$, 

$f = \sum m(1, 6, 8, 15)$, 

then $f_2$ is 

(A) $\sum m(4, 6)$ 

(B) $\sum m(4, 8)$ 

(C) $\sum m(6, 8)$ 

(D) $\sum m(4, 6, 8)$ 

(GATE 2008) 

Answer: (C) $\sum m(6, 8)$ 

Explanantion: 

$f = (f_1 \cap f_2) \cup f_3$ 

Check option (C): 

$(f_1 \cap f_2) \cup f_3$ 

$= \big(\sum m(4, 5, 6, 7, 8) \cap \sum m(6, 8)\big) \cup \sum m(1, 6, 15)$ 

$= \sum m(6, 8) \cup \sum m(1, 6, 15)$ 

$= \sum m(1, 6, 8, 15)$  

$= f$ 

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