The given logic circuit consists of an OR gate and a NAND gate.
The output of the OR gate is y=n+m.
The inputs to the NAND gate are n and y.
The final output z is given by z=n⋅y=n⋅(n+m).
Using Boolean algebra: n⋅(n+m)=n⋅n+n⋅m=n+n⋅m=n(1+m)=n.
Therefore, z=n.
Let's verify this with the truth table:
| n | m | y = n + m | z = NAND(n, y) |
|---|---|---|---|
| 0 | 0 | 0 | 1 |
| 0 | 1 | 1 | 1 |
| 1 | 1 | 1 | 0 |
| 1 | 0 | 1 | 0 |
Comparing this with the given options, option (4) matches the calculated truth table.