Question

$(p\mspace{6mu} \land \sim q) \land (\sim p \land q)$ is

• A. A tautology
• B. A contradiction
• C. Both a tautology and a contradiction
• D. Neither a tautology nor a contradiction

Answer 1

Answer: (B)

A contradiction

Answer 2

$(p\mspace{6mu} \land \sim q) \land (\sim p \land q) = (p\mspace{6mu} \land \sim p) \land (\sim q \land q) = f \land f = f$.

(By using associative laws and commutatine laws)

∴ $(p \land \sim q) \land (\sim p \land q)$ is a contradiction.

($(p\mspace{6mu} \land \sim q) \land (\sim p \land q) = (p\mspace{6mu} \land \sim p) \land (\sim q \land q) = f \land f = f$.