Let $$\Delta \in \{\wedge, \vee, \Rightarrow, \Leftrightarrow\}$$ be such that $$(p \wedge q)\Delta((p \vee q) \Rightarrow q)$$ is a tautology. Then $$\Delta$$ is equal to

We need to find $$\Delta \in \{\wedge, \vee, \Rightarrow, \Leftrightarrow\}$$ such that $$(p \wedge q) \;\Delta\; ((p \vee q) \Rightarrow q)$$ is a tautology.

Evaluate both components for all truth value combinations:

Let $$X = p \wedge q$$ and $$Y = (p \vee q) \Rightarrow q$$.

Case 1: $$p = T, q = T$$: $$X = T$$, $$p \vee q = T$$, $$Y = T \Rightarrow T = T$$. So $$T \;\Delta\; T$$.

Case 2: $$p = T, q = F$$: $$X = F$$, $$p \vee q = T$$, $$Y = T \Rightarrow F = F$$. So $$F \;\Delta\; F$$.

Case 3: $$p = F, q = T$$: $$X = F$$, $$p \vee q = T$$, $$Y = T \Rightarrow T = T$$. So $$F \;\Delta\; T$$.

Case 4: $$p = F, q = F$$: $$X = F$$, $$p \vee q = F$$, $$Y = F \Rightarrow F = T$$. So $$F \;\Delta\; T$$.

For a tautology, all four cases must give True:

From Case 1: $$T \;\Delta\; T = T$$ — true for all four connectives.

From Case 2: $$F \;\Delta\; F = T$$ — checking: $$\wedge$$: F, $$\vee$$: F, $$\Rightarrow$$: T, $$\Leftrightarrow$$: T. Only $$\Rightarrow$$ and $$\Leftrightarrow$$ work.

From Cases 3 and 4: $$F \;\Delta\; T = T$$ — checking: $$\Rightarrow$$: T, $$\Leftrightarrow$$: F. Only $$\Rightarrow$$ works.

Only $$\Delta = \Rightarrow$$ satisfies all cases.

The answer is Option C: $$\Rightarrow$$.