Let the operations $$*, \odot \in \{\wedge, \vee\}$$. If $$(p * q) \odot (p \odot \sim q)$$ is a tautology, then the ordered pair $$(*, \odot)$$ is

We need to find the ordered pair $$(*, \odot)$$ from $$\{\wedge, \vee\}$$ such that $$(p * q) \odot (p \odot \sim q)$$ is a tautology.

Step 1: Test each option systematically.

Option B: $$(*, \odot) = (\vee, \vee)$$

The expression becomes: $$(p \vee q) \vee (p \vee \sim q)$$

$$= p \vee q \vee p \vee \sim q$$

$$= p \vee (q \vee \sim q)$$

$$= p \vee T$$

$$= T$$

This is a tautology.

Step 2: Verify that other options are not tautologies.

Option A: $$(*, \odot) = (\vee, \wedge)$$

$$(p \vee q) \wedge (p \wedge \sim q)$$

When $$p = F, q = T$$: $$(F \vee T) \wedge (F \wedge F) = T \wedge F = F$$. Not a tautology.

Option C: $$(*, \odot) = (\wedge, \wedge)$$

$$(p \wedge q) \wedge (p \wedge \sim q)$$

$$= p \wedge (q \wedge \sim q) = p \wedge F = F$$. Not a tautology (always false).

Option D: $$(*, \odot) = (\wedge, \vee)$$

$$(p \wedge q) \vee (p \vee \sim q)$$

When $$p = F, q = T$$: $$(F \wedge T) \vee (F \vee F) = F \vee F = F$$. Not a tautology.

Only Option B gives a tautology.

The correct answer is Option B: $$(\vee, \vee)$$