GATE CS 2026 Set 2 Q.11: Propositional and First-Order Logic

For two different persons $x$ and $y$ , the predicate $M(x, y)$ denotes that $x$ knows $y$ . Consider the following statement.

There is a person who does not know anyone else, but that person is known by everyone else.

Which one of the following expressions represents the above statement?

$(\exists y)(\forall x) ((x \neq y) \rightarrow (M(x, y) \wedge \neg M(y, x)))$

$(\forall y)(\exists x) ((x \neq y) \rightarrow (M(x, y) \wedge \neg M(y, x)))$

$(\exists y)(\exists x) ((x \neq y) \rightarrow (M(x, y) \wedge \neg M(y, x)))$

$(\forall y)(\forall x) ((x \neq y) \rightarrow (M(x, y) \wedge \neg M(y, x)))$

Official answer and solution

Official answer: A
From: Official GATE 2026 answer key

Solution

Bind the existential quantifier $\exists y$ to 'there is a person' and the universal quantifier $\forall x$ to 'everyone else'.

Let $y$ be the special person ('there is a person': $\exists y$ ). For every other person $x$ ( $\forall x$ with $x \neq y$ ), two conditions hold: $x$ knows $y$ ( $M(x, y)$ ) and $y$ does not know $x$ ( $\neg M(y, x)$ ). Combining these yields:

(\exists y)(\forall x) ((x \neq y) \rightarrow (M(x, y) \wedge \neg M(y, x)))

Final answer

Option (A)

Common trap

Reversing the quantifier order to $\forall y \exists x$ .

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2026 Computer Science and Information Technology - Session 2 (CS2)
Question: 11 of 65
Section: Computer Science and Information Technology

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