Question

For p and q, consider: (a) (~ q ∧ (p → q)) → ~ p, (b) ((p ∨ q) ∧ ~ p) → q. Which is correct ?

• (a) is a tautology but not (b).
• (b) is a tautology but not (a).
• (a) and (b) both are tautologies.
• (a) and (b) both are not tautologies.

Hint:

The logical form "If (A and B) then A" is always a tautology. Use distribution laws to simplify the antecedent.

Answer 1

The Correct Option is C

Step 1: For (a): $p \to q \equiv \sim p \vee q$. So $\sim q \wedge (\sim p \vee q) \equiv (\sim q \wedge \sim p) \vee (\sim q \wedge q) \equiv \sim q \wedge \sim p$. The statement becomes $(\sim q \wedge \sim p) \to \sim p$. This is of the form $(X \wedge Y) \to Y$, which is always true (Tautology).
Step 2: For (b): $(p \vee q) \wedge \sim p \equiv (p \wedge \sim p) \vee (q \wedge \sim p) \equiv F \vee (q \wedge \sim p) \equiv q \wedge \sim p$. The statement becomes $(q \wedge \sim p) \to q$. This is also $(X \wedge Y) \to X$, always true (Tautology).