If the Boolean expression $$(p \Rightarrow q) \Leftrightarrow (q * (\sim p))$$ is a tautology, then the Boolean expression $$p * (\sim q)$$ is equivalent to:

We are given that $$(p \Rightarrow q) \Leftrightarrow (q * (\sim p))$$ is a tautology, and we need to find what $$p * (\sim q)$$ is equivalent to.

Recall the standard logical equivalence: $$p \Rightarrow q \equiv \sim p \lor q$$. This means "if p then q" is the same as "not-p or q".

For the biconditional $$(p \Rightarrow q) \Leftrightarrow (q * (\sim p))$$ to be a tautology, both sides must always have the same truth value. So we need $$\sim p \lor q \equiv q * (\sim p)$$ for every combination of truth values of $$p$$ and $$q$$.

Let us substitute $$a = \sim p$$ to simplify. Then we need $$a \lor q \equiv q * a$$ for all truth values of $$a$$ and $$q$$. Since $$a$$ and $$q$$ range over all truth values independently (as $$p$$ and $$q$$ do), this means the operation $$*$$ must be exactly the same as $$\lor$$ (OR).

To verify with a truth table: when $$a = T, q = T$$: $$a \lor q = T$$ so $$q * a = T$$. When $$a = T, q = F$$: $$a \lor q = T$$ so $$q * a = T$$. When $$a = F, q = T$$: $$a \lor q = T$$ so $$q * a = T$$. When $$a = F, q = F$$: $$a \lor q = F$$ so $$q * a = F$$. This matches the OR truth table exactly.

Now we evaluate $$p * (\sim q)$$. Since $$*$$ is $$\lor$$, we get $$p * (\sim q) = p \lor (\sim q)$$.

Using the equivalence $$q \Rightarrow p \equiv \sim q \lor p = p \lor (\sim q)$$, we see that $$p * (\sim q) \equiv q \Rightarrow p$$.

This matches Option A: $$q \Rightarrow p$$.