Let $p$, $q$, and $r$ be propositions and the expression $(p \to q) \to r$ be a contradiction. Then, the expression $(r \to p) \to q$ is

If the expression $(p \to q) \to r$ is a contradiction (always FALSE), then for the conditional to be FALSE:

1. The antecedent $(p \to q)$ must be TRUE.

2. The consequent $r$ must be FALSE.

Now consider the expression $(r \to p) \to q$:

• Since $r$ is FALSE, the implication $(r \to p)$ is vacuously TRUE regardless of the truth value of $p$.

• Substituting $(r \to p) = \text{TRUE}$ into the expression gives:

Thus, the expression $(r \to p) \to q$ has the same truth value as $q$. Therefore, it is always TRUE when $q$ is TRUE.

Hence, the correct option is (D).