Which of the following statement is a tautology?

To decide which given statement is a tautology, we consider every option one by one and simplify it algebraically.

First, recall the basic logical equivalence for an implication:

$$a \rightarrow b \;\equiv\; \sim a \;\vee\; b.$$

We will convert each implication using this rule and then simplify.

Option A is $$p \,\vee\, (\sim q) \;\rightarrow\; p \,\wedge\, q.$$

Using the implication rule we have

$$\sim\bigl(p \,\vee\, (\sim q)\bigr) \;\vee\; (p \,\wedge\, q).$$

Apply De Morgan’s law to the negation:

$$\sim p \;\wedge\; q \;\vee\; p \;\wedge\; q.$$

Now take $$q$$ common:

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

Inside the parentheses, $$\sim p \;\vee\; p$$ is a tautology (it is always true). Therefore the whole expression reduces to

$$q \;\wedge\; \text{(True)} \;=\; q.$$

This final result $$q$$ is not always true (it is false when $$q$$ is false), so Option A is not a tautology.

Option B is $$\sim(p \,\wedge\, \sim q) \;\rightarrow\; p \,\vee\, q.$$

Convert the implication:

$$\sim\!\bigl(\sim(p \,\wedge\, \sim q)\bigr) \;\vee\; (p \,\vee\, q).$$

The double negation simplifies immediately:

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

Group the $$p$$ terms:

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

Using the absorption law $$p \,\vee\, (p \,\wedge\, r)=p,$$ we obtain

$$p \;\vee\; q.$$

The expression $$p \;\vee\; q$$ can be false (specifically when $$p$$ and $$q$$ are both false). Thus Option B is not a tautology.

Option C is $$\sim(p \,\vee\, \sim q) \;\rightarrow\; p \,\wedge\, q.$$

Again convert the implication:

$$\sim\!\bigl(\sim(p \,\vee\, \sim q)\bigr) \;\vee\; (p \,\wedge\, q).$$

The double negation gives

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

Take $$p$$ common from the last two pieces:

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

Using absorption again, $$p \;\vee\; (p \,\wedge\, q)=p$$, so we get

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

This can be false whenever $$p$$ is false and $$q$$ is true. Hence Option C is not a tautology.

Option D is $$\sim(p \,\vee\, \sim q) \;\rightarrow\; p \,\vee\, q.$$

Convert the implication exactly as before:

$$\sim\!\bigl(\sim(p \,\vee\, \sim q)\bigr) \;\vee\; (p \,\vee\, q).$$

Simplify the double negation:

$$(p \,\vee\, \sim q) \;\vee\; (p \,\vee\, q).$$

Rearrange the terms:

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

Using idempotent law $$p \,\vee\, p = p,$$ merge the repeated $$p$$ terms:

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

Now notice that $$\sim q \;\vee\; q$$ is always true (law of excluded middle), so write

$$p \;\vee\; \text{(True)}.$$

Finally, $$\alpha \;\vee\; \text{(True)} = \text{True}$$ for any statement $$\alpha$$. Therefore the whole expression is always true, no matter what truth-values $$p$$ and $$q$$ take.

Thus Option D represents a statement that is true in every possible case; that is exactly the definition of a tautology.

Hence, the correct answer is Option D.