Which of the following statements is a tautology?

To determine which statement is a tautology (always true regardless of truth values), we evaluate each option using truth tables or logical equivalences. We consider all combinations of p and q (True/False).

Option A: $$p \to (p \wedge (p \to q))$$

Compute step by step:

Case 1: p=T, q=T
$$p \to q = T \to T = T$$
$$p \wedge (p \to q) = T \wedge T = T$$
$$p \to T = T \to T = T$$

Case 2: p=T, q=F
$$p \to q = T \to F = F$$
$$p \wedge F = T \wedge F = F$$
$$p \to F = T \to F = F$$ → Not always true

Thus, option A is not a tautology.

Option B: $$(p \wedge q) \to (\sim p \to q)$$

Use logical equivalence: $$A \to B \equiv \sim A \vee B$$

First, simplify the inner implication:
$$\sim p \to q \equiv \sim(\sim p) \vee q \equiv p \vee q$$

So the expression becomes:
$$(p \wedge q) \to (p \vee q)$$
Apply equivalence again:
$$\sim(p \wedge q) \vee (p \vee q)$$
By De Morgan's law: $$\sim(p \wedge q) \equiv \sim p \vee \sim q$$
Thus: $$(\sim p \vee \sim q) \vee (p \vee q)$$
Rearrange using associativity and commutativity:
$$(\sim p \vee p) \vee (\sim q \vee q)$$
Now, $$\sim p \vee p \equiv T$$ (tautology) and $$\sim q \vee q \equiv T$$
So: $$T \vee T \equiv T$$ → Always true

Thus, option B is a tautology.

Option C: $$(p \wedge (p \to q)) \to \sim q$$

Case 1: p=T, q=T
$$p \to q = T \to T = T$$
$$p \wedge T = T \wedge T = T$$
$$\sim q = F$$
$$T \to F = F$$ → Not always true

Thus, option C is not a tautology.

Option D: $$p \vee (p \wedge q)$$

Case 1: p=T, q=T → $$T \vee (T \wedge T) = T \vee T = T$$
Case 2: p=T, q=F → $$T \vee (T \wedge F) = T \vee F = T$$
Case 3: p=F, q=T → $$F \vee (F \wedge T) = F \vee F = F$$ → Not always true
Case 4: p=F, q=F → $$F \vee (F \wedge F) = F \vee F = F$$

Thus, option D is not a tautology.

Only option B is always true. The correct answer is option B (the second option).