Official PYQ · 2015Set 2 · Q37Boolean AlgebraNAT1 mark

GATE CS 2015 Set 2 Q37 — Boolean Algebra

The number of min-terms after minimizing the following Boolean expression is _______.

[D'+AB'+A'C+AC'D+A'C'D]'

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Answer: 1

To find the number of min-terms of the given Boolean expression, let us first simplify the expression inside the complement.

Let the given expression be $F = G'$ , where:

G = D' + AB' + A'C + AC'D + A'C'D

We can group the terms containing $C'D$ :

AC'D + A'C'D = (A + A')C'D = C'D

Substitute this back into the expression for $G$ :

G = D' + AB' + A'C + C'D

Using the distributive law on $D' + C'D$ :

D' + C'D = (D' + C')(D' + D) = D' + C'

Now, substitute this back into $G$ :

G = D' + C' + AB' + A'C

Using the identity $X + X'Y = X + Y$ on $C' + A'C$ (with $X = C'$ and $Y = A'$ ):

C' + A'C = C' + A'

This simplifies $G$ to:

G = D' + C' + A' + AB'

Similarly, using the identity $X' + XY = X' + Y$ on $A' + AB'$ (with $X = A$ and $Y = B'$ ):

A' + AB' = A' + B'

Thus, the fully simplified expression for $G$ is:

G = A' + B' + C' + D'

Using De Morgan's Law:

F = (A' + B' + C' + D')' = A \cdot B \cdot C \cdot D

The minimized expression is $F = ABCD$ , which represents exactly 1 min-term ( $m_{15}$ ).

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Published 10 Oct 2026

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