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Official PYQ · 2002Q2-3Boolean AlgebraMCQ1 mark

GATE CS 2002 Q2-3 — Boolean Algebra

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Let f(A,B)=A′+Bf(A,B) = A'+B. Simplified expression for function f(f(x+y,y),z)f(f(x+y, y), z) is

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Answer: C

We are given the function definition:

f(A,B)=A′+Bf(A, B) = A' + B

We need to find the simplified expression for f(f(x+y,y),z)f(f(x+y, y), z). Let us evaluate this step-by-step.

Step 1: Evaluate the inner function f(x+y,y)f(x+y, y)

Substitute A=x+yA = x+y and B=yB = y into the function definition:

f(x+y,y)=(x+y)′+y=x′y′+y(by De Morgan’s Law)=(x′+y)(y′+y)(by Distributive Law)=(x′+y)(1)=x′+y\begin{aligned} f(x+y, y) &= (x+y)' + y \\ &= x'y' + y & \quad \text{(by De Morgan's Law)} \\ &= (x' + y)(y' + y) & \quad \text{(by Distributive Law)} \\ &= (x' + y)(1) \\ &= x' + y \end{aligned}

Step 2: Evaluate the outer function f(f(x+y,y),z)f(f(x+y, y), z)

Substitute the result from Step 1, f(x+y,y)=x′+yf(x+y, y) = x' + y, into the outer function:

f(x′+y,z)=(x′+y)′+z=(x′)′⋅y′+z(by De Morgan’s Law)=xy′+z\begin{aligned} f(x'+y, z) &= (x' + y)' + z \\ &= (x')' \cdot y' + z & \quad \text{(by De Morgan's Law)} \\ &= xy' + z \end{aligned}

Thus, the simplified expression is xy′+zxy' + z, which corresponds to option C.

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