Let $$\Delta, \nabla \in \{\wedge, \vee\}$$ be such that $$p\nabla q \to ((p\Delta q)\nabla r)$$ is a tautology. Then $$(p\nabla q) \Delta r$$ is logically equivalent to

We need to find $$\Delta, \nabla \in \{\wedge, \vee\}$$ such that $$p \nabla q \to ((p \Delta q) \nabla r)$$ is a tautology, then determine what $$(p \nabla q) \Delta r$$ is logically equivalent to.

Since $$\nabla$$ and $$\Delta$$ can each be $$\vee$$ or $$\wedge$$, there are four possibilities. First, if $$\nabla = \vee$$ and $$\Delta = \vee$$, then $$p \vee q \to (p \vee q) \vee r$$. This is always true since $$(p \vee q) \vee r$$ contains $$p \vee q$$, so it is a tautology. Next, if $$\nabla = \vee$$ and $$\Delta = \wedge$$, then $$p \vee q \to (p \wedge q) \vee r$$. Taking $$p = T, q = F, r = F$$ yields $$T \to F$$, which is false, so this case is not a tautology. Now if $$\nabla = \wedge$$ and $$\Delta = \vee$$, then $$p \wedge q \to (p \vee q) \wedge r$$. Taking $$p = T, q = T, r = F$$ gives $$T \to F$$, which is false. Similarly, if $$\nabla = \wedge$$ and $$\Delta = \wedge$$, then $$p \wedge q \to (p \wedge q) \wedge r$$, and choosing the same values yields $$T \to F$$ again. Therefore the only tautology arises when $$\nabla = \vee$$ and $$\Delta = \vee$$.

Substituting these back shows that $$(p \nabla q) \Delta r$$ becomes $$(p \vee q) \vee r$$, which simplifies to $$p \vee q \vee r$$.

Among the answer choices, Option A: $$(p \Delta r) \vee q = (p \vee r) \vee q = p \vee q \vee r$$ matches this expression. The correct answer is Option A.

The correct answer is Option A.