Which of the following is equivalent to the Boolean expression $$p \wedge \sim q$$?

We have to find an option whose logical value is always identical to the expression $$p \wedge \sim q$$. The expression $$p \wedge \sim q$$ is true exactly when $$p$$ is true and $$q$$ is false, and false in every other situation. We shall translate each option into a form that can be compared directly with this conjunction.

First, we recall and state the fundamental implication law: for any statements $$x$$ and $$y$$,

$$x \rightarrow y \;\equiv\; \sim x \;\vee\; y.$$

Using this single law and ordinary De Morgan simplifications $$\sim(x \vee y)=\sim x \wedge \sim y$$ and $$\sim(x \wedge y)=\sim x \vee \sim y,$$ we convert every option step by step.

Option A is $$\sim p \rightarrow \sim q$$. Applying the implication law with $$x=\sim p$$ and $$y=\sim q$$ gives

$$\sim p \rightarrow \sim q \;\equiv\; \sim(\sim p) \;\vee\; \sim q.$$

Now $$\sim(\sim p)=p$$, so

$$\sim p \rightarrow \sim q \;\equiv\; p \;\vee\; \sim q.$$

Thus Option A simplifies to $$p \vee \sim q$$, a disjunction, not the required conjunction $$p \wedge \sim q$$. Hence Option A is not equivalent.

Option B is $$\sim\bigl(q \rightarrow p\bigr)$$. First write the inner implication:

$$q \rightarrow p \;\equiv\; \sim q \;\vee\; p.$$

Substituting, we get

$$\sim\bigl(q \rightarrow p\bigr) \;=\; \sim\bigl(\sim q \;\vee\; p\bigr).$$

Using De Morgan’s law for negation of a disjunction,

$$\sim(\sim q \;\vee\; p) \;=\; \sim(\sim q) \;\wedge\; \sim p.$$

Now $$\sim(\sim q)=q,$$ so

$$\sim\bigl(q \rightarrow p\bigr) \;\equiv\; q \;\wedge\; \sim p.$$

This is the conjunction $$q \wedge \sim p,$$ which is different from $$p \wedge \sim q$$. Therefore Option B is also not equivalent.

Option C is $$\sim\bigl(p \rightarrow q\bigr).$$ Again we expand the implication:

$$p \rightarrow q \;\equiv\; \sim p \;\vee\; q.$$

Substituting gives

$$\sim\bigl(p \rightarrow q\bigr) \;=\; \sim\bigl(\sim p \;\vee\; q\bigr).$$

Using De Morgan’s law,

$$\sim(\sim p \;\vee\; q) \;=\; \sim(\sim p) \;\wedge\; \sim q.$$

Simplifying $$\sim(\sim p)=p,$$ we obtain

$$\sim\bigl(p \rightarrow q\bigr) \;\equiv\; p \;\wedge\; \sim q.$$

This is exactly the original expression we wanted. So Option C matches $$p \wedge \sim q$$ perfectly.

Option D is $$\sim\bigl(p \rightarrow \sim q\bigr).$$ Start with the implication:

$$p \rightarrow \sim q \;\equiv\; \sim p \;\vee\; \sim q.$$

Negating it,

$$\sim\bigl(p \rightarrow \sim q\bigr) \;=\; \sim\bigl(\sim p \;\vee\; \sim q\bigr).$$

Applying De Morgan’s law,

$$\sim(\sim p \;\vee\; \sim q) \;=\; \sim(\sim p) \;\wedge\; \sim(\sim q).$$

This simplifies to

$$p \;\wedge\; q.$$

Since $$p \wedge q$$ differs from $$p \wedge \sim q,$$ Option D is not equivalent.

Among all four options, only Option C reduces to $$p \wedge \sim q$$. Hence, the correct answer is Option 3.