Q.The negation of inverse of $\sim p \to q$ is ________.
A) $q \wedge p$
B) $\sim p \wedge \sim q$
C) $p \vee q$
D) $\sim q \to \sim p$

Find the contrapositive of $\sim p\rightarrow q$, then take its converse.

The contrapositive of a conditional $a\rightarrow b$ is $\sim b\rightarrow \sim a$.

Here $a=\sim p,\ b=q$. So the contrapositive of $\sim p\rightarrow q$ is:

$\sim q\rightarrow \sim(\sim p) \equiv \sim q\rightarrow p$

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With $p:x<y,\ q:x^2<y^2$, inverse is $\sim p\to\sim q$, contrapositive is $\sim q\to\sim p$.

Let $p: x<y$ and $q: x^2<y^2$, so the statement is $p\to q$.

Inverse ($\sim p\to\sim q$): If $x\ge y$ then $x^2\ge y^2$.

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