The statement $$p \to q \to (\sim p \to q \to q)$$ is

We have the propositional statement

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

The implication sign $$\rightarrow$$ is taken to be right-associative, so an expression like $$a \rightarrow b \rightarrow c$$ is interpreted as $$a \rightarrow (\,b \rightarrow c\,).$$ Using this convention, our statement can be rewritten (placing all invisible parentheses explicitly) as

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

Now we proceed from the innermost part and gradually simplify.

First, consider the sub-statement $$q \;\rightarrow\; q.$$

• By the truth table for implication, a statement of the form $$A \rightarrow A$$ is always true, because:   • If $$A$$ is true, then $$A \rightarrow A$$ has true antecedent and true consequent, so it is true.
  • If $$A$$ is false, then $$A \rightarrow A$$ has false antecedent, and an implication with a false antecedent is true.
Hence

$$q \;\rightarrow\; q \equiv \text{T},$$

where “T” denotes the truth constant “always true”.

Substituting this result into the expression, we obtain

$$p \;\rightarrow\; \bigl(\,q \;\rightarrow\; \bigl(\,(\sim p) \;\rightarrow\; \text{T}\bigr)\bigr).$$

Next, look at the part $$(\sim p) \;\rightarrow\; \text{T}.$$

• A basic property of implication is: a statement of the form $$A \rightarrow \text{T}$$ is always true, because its consequent is true regardless of $$A$$. Therefore

$$(\sim p) \;\rightarrow\; \text{T} \equiv \text{T}.$$

Replacing this with “T”, the larger expression simplifies to

$$p \;\rightarrow\; \bigl(\,q \;\rightarrow\; \text{T}\bigr).$$

Again apply the same rule to $$q \;\rightarrow\; \text{T}$$: since the consequent is “T”, we have

$$q \;\rightarrow\; \text{T} \equiv \text{T}.$$

This leaves the outermost implication:

$$p \;\rightarrow\; \text{T}.$$

And finally, by the very same principle, any implication whose consequent is “T” is itself always true, so

$$p \;\rightarrow\; \text{T} \equiv \text{T}.$$

Thus every possible truth-value assignment to $$p$$ and $$q$$ makes the given statement true. Therefore the original statement is a tautology.

Hence, the correct answer is Option A.