Logic gateHard

Question

For the given logic gate circuit, which of the following is the correct truth table?

Options

A.
B.
C.
D.

Solution

Gate 1: At bottom there is an OR gate with inputs n & m

out put $= n + m$

Gate 2 : A NAND gate, its Input are direct n & the output of OR gate ( $n + m$ )

out put $z = \overline{n \cdot (n + m)}$

since $n.(n + m) = (n.n) + (n.m)$

$$= n + n.m = n(1 + m) = n $$∴ output $z = \overline{n \cdot (n + m)} = \overline{n}$

n m $$z = \overline{n}$$
0 0 1
0 1 1
1 1 0
1 0 0

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