JEE Main · 2022 · 26 Jun 2022 · Shift 1 · Mathematical Reasoning *

Testing the cases:

Case I: If $\Delta \equiv \nabla \equiv \wedge$, then $(p\wedge q)\to ((p\wedge q)\wedge r)$ is not a tautology (false when r is false).

Case II: If $\Delta \equiv \nabla \equiv \vee$, then $(p\vee q)\to ((p\vee q)\vee r)$ is a tautology. In this case, $(p\vee q)\vee r\equiv (p\Delta r)\vee q$.

Case III: If $\Delta =\vee ,\nabla =\wedge$, then $(p\wedge q)\to \{(p\vee q)\wedge r\}$ is not a tautology.

Case IV: If $\Delta =\wedge ,\nabla =\vee$, then $(p\vee q)\to \{(p\wedge q)\vee r\}$ is not a tautology.

Therefore, $\Delta \equiv \nabla \equiv \vee$ and the answer is $(p\Delta r)\vee q$.