The negation of the Boolean expression x- y is equivalent to

We know that, p↔ q=(p→ q)(q→ p) ∼(p↔ q)=(p ∼ q)(∼ p q) Now, replacing q with ∼ q, we have ∼(p↔∼ q)=(p q)(∼ p ∼ q) ∴ ∼(x↔∼ y)=(x y) (∼ x ∼ y) nbsp; ...

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Standard XII

Mathematics

Distributive Law

The negation ...

Question

The negation of the Boolean expression $x \leftrightarrow\sim y$ is equivalent to

(x \land y) \land (\sim x \lor \sim y)

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(x \land y) \lor (\sim x \land \sim y)

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(x \land \sim y) \lor (\sim x \land y)

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(\sim x \land y) \lor (\sim x \land \sim y)

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Solution

The correct option is B $(x \land y) \lor (\sim x \land \sim y)$
We know that, $p \leftrightarrow q = (p \to q) \land (q \to p)$
$\sim (p \leftrightarrow q) = (p \land \sim q) \lor (\sim p \land q)$
Now, replacing $q$ with $\sim q$ , we have
$\sim (p \leftrightarrow\sim q) = (p \land q) \lor (\sim p \land \sim q)$

$∴ \sim (x \leftrightarrow\sim y) = (x \land y) \lor (\sim x \land \sim y)$

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The negation of the Boolean expression x↔∼y is equivalent to

The correct option is B (x∧y)∨(∼x ∧∼y) We know that, p↔q=(p→q)∧(q→p) ∼(p↔q)=(p ∧∼q)∨(∼p∧q) Now, replacing q with ∼q, we have ∼(p↔∼q)=(p∧q)∨(∼p ∧∼q) ∴ ∼(x↔∼y)=(x∧y)∨(∼x ∧∼y)

The negation of the Boolean expression $x \leftrightarrow\sim y$ is equivalent to