NIMCET 2026 — Computer PYQ

To simplify the given Product-of-Sums (POS) Boolean expression, we can apply standard Boolean algebraic laws.

Given Expression:

$$E=(x+y^{\prime }+z^{\prime })(x+y^{\prime }+z)(x+y+z^{\prime })$$

Step 1: Group and simplify the first two terms

Look at the first two factors: $(x+y^{\prime }+z^{\prime })$ and $(x+y^{\prime }+z)$.

Notice that $(x+y^{\prime })$ is common to both terms. We can use the Distributive Law $A+BC=(A+B)(A+C)$, where we let $A=(x+y^{\prime })$, $B=z^{\prime }$, and $C=z$:

$$(x+y^{\prime }+z^{\prime })(x+y^{\prime }+z)=(x+y^{\prime })+(z^{\prime }\cdot z)$$

Since $z^{\prime }\cdot z=0$ (Complement Law):

$$(x+y^{\prime })+0=x+y^{\prime }$$

Now, substitute this back into the main expression.

Step 2: Multiply with the third term

Our expression now simplifies to:

$$E=(x+y^{\prime })(x+y+z^{\prime })$$

We can apply the Distributive Law again. Notice that $x$ is common to both terms ($x+A)(x+B)=x+AB$:

$$E=x+[y^{\prime }\cdot (y+z^{\prime })]$$

Step 3: Expand and simplify the remaining part

Now, distribute $y^{\prime }$ inside the brackets:

$$E=x+(y^{\prime }\cdot y+y^{\prime }\cdot z^{\prime })$$

Since $y^{\prime }\cdot y=0$ (Complement Law):

$$E=x+(0+y^{\prime }{z}^{\prime })$$

$$E=x+y^{\prime }{z}^{\prime }$$

Conclusion

The simplified Boolean expression matches Option 2.

$$\text{Final Answer: }x+y^{\prime }{z}^{\prime }$$