Negation of \(\mathrm{p} \rightarrow \mathrm{q}\) is (a) \(\sim \mathrm{p} \rightarrow \sim \mathrm{q}\) (b) \(\mathrm{p} \rightarrow \sim \mathrm{q}\) (c) \(\mathrm{p} \rightarrow \mathrm{q}\) (d) \(\sim \mathrm{q} \rightarrow \sim \mathrm{p}\)

In propositional logic, logical connectives are used to link different propositions/statement variables. Here we have two connectives: 1. Implication (denoted by \(\rightarrow\)): the implication \(\mathrm{p} \rightarrow \mathrm{q}\) means 'if p, then q' or 'p implies q'. This connective is true unless p is true and q is false. 2. Negation (denoted by \(\sim\)): the negation \(\sim \mathrm{p}\) means 'not p'. This connective simply reverses the truth value of the proposition it's applied to.

Recall that an implication can also be written as a disjunction (logical OR, denoted by \(\vee\)) along with negation, as follows: \(\mathrm{p} \rightarrow \mathrm{q} \equiv \sim \mathrm{p} \vee \mathrm{q}\) This means that the implication \(\mathrm{p} \rightarrow \mathrm{q}\) is logically equivalent to the statement 'either p is false or q is true'.

Now, we will negate the equivalent expression of the implication \(\sim \mathrm{p} \vee \mathrm{q}\), using the De Morgan's law, which states that: \(\sim (\mathrm{A} \vee \mathrm{B}) \equiv (\sim \mathrm{A} \wedge \sim \mathrm{B})\) \(\sim (\sim \mathrm{p} \vee \mathrm{q}) \equiv (\sim (\sim \mathrm{p}) \wedge \sim \mathrm{q})\) Applying double negation (as \(\sim(\sim \mathrm{A})\) is the same as \(\mathrm{A}\)): \(\sim (\sim \mathrm{p} \vee \mathrm{q}) \equiv (\mathrm{p} \wedge \sim \mathrm{q})\)

Finally, let's compare the negated expression \((\mathrm{p} \wedge \sim \mathrm{q})\) with the provided options: (a) \(\sim \mathrm{p} \rightarrow \sim \mathrm{q}\) (b) \(\mathrm{p} \rightarrow \sim \mathrm{q}\) (c) \(\mathrm{p} \rightarrow \mathrm{q}\) (d) \(\sim \mathrm{q} \rightarrow \sim \mathrm{p}\) None of these options exactly match our negated expression \((\mathrm{p} \wedge \sim \mathrm{q})\). However, (b) is the closest because it also states that \(\mathrm{p}\) is true and \(\mathrm{q}\) is false, though using implication notation. It's important to note that (b) isn't equivalent to the negated expression but is the closest option provided. Thus, we can choose the closest answer, which is - (b) \(\mathrm{p} \rightarrow \sim \mathrm{q}\).