The statement $$(p \wedge (p \rightarrow q) \wedge (q \rightarrow r)) \rightarrow r$$ is

We have to inspect the compound statement $$\bigl(p \wedge (p \rightarrow q) \wedge (q \rightarrow r)\bigr) \rightarrow r$$ and decide whether it is always true (a tautology), always false (a fallacy), or equivalent to some simpler implication.

First of all, recall the standard equivalence for an implication:

$$x \rightarrow y \;\;\text{is logically equal to}\;\; \sim x \vee y.$$

Using this rule, we rewrite each implication inside the larger expression.

$$p \rightarrow q \equiv \sim p \vee q,$$

$$q \rightarrow r \equiv \sim q \vee r.$$

Substituting these into the original statement gives

$$\Bigl(p \wedge (\sim p \vee q) \wedge (\sim q \vee r)\Bigr) \rightarrow r.$$

Let us now simplify the conjunction in the antecedent, step by step.

First combine the first two factors:

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

But $$p \wedge \sim p$$ is a contradiction, hence equal to false. So

$$(p \wedge \sim p) \vee (p \wedge q) = \text{false} \vee (p \wedge q) = p \wedge q.$$

Thus the antecedent has now become

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

We keep going. Group the last two factors:

$$q \wedge (\sim q \vee r) = (q \wedge \sim q) \vee (q \wedge r).$$

Again $$q \wedge \sim q$$ is a contradiction, so this reduces to

$$(q \wedge \sim q) \vee (q \wedge r) = \text{false} \vee (q \wedge r) = q \wedge r.$$

Hence the whole antecedent simplifies neatly to

$$p \wedge (q \wedge r) = p \wedge q \wedge r.$$

The complete statement is therefore

$$(p \wedge q \wedge r) \rightarrow r.$$

Apply the implication equivalence once more:

$$(p \wedge q \wedge r) \rightarrow r \equiv \sim(p \wedge q \wedge r) \vee r.$$

By De Morgan’s law,

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

So the entire disjunction becomes

$$\bigl(\sim p \vee \sim q \vee \sim r\bigr) \vee r.$$

Now observe that $$\sim r \vee r$$ is a tautology (it is always true). Because that tautology is one of the disjuncts, the whole expression is invariably true, no matter what truth-values $$p, q,$$ and $$r$$ may take.

Therefore the original statement is always true; that is, it is a tautology.

Hence, the correct answer is Option A.