(p ∧ ∼ q) ∧ ( ∼ p ∧ q) is

Question

(p ∧ ∼ q) ∧ ( ∼ p ∧ q) is (A) tautology (B) a contradiction (C) both a tautology and a contradiction (D) neither a tautology nor a contradiction

Tardigrade Admin · Accepted Answer

( p ∧ ∼ q) ∧ ( ∼ p ∧ q) = ( p ∧ ∼ p) ∧ ( ∼ q ∧ q) = f ∧ f = f (By using associative laws and commutative laws). ∴ (p ∧ ∼ q) ∧ ( ∼ p ∧ q) is a contradiction ∴ a contradiction holds

Q. $(p \land \sim q) \land (\sim p \land q)$ is

tautology

a contradiction

both a tautology and a contradiction

neither a tautology nor a contradiction

$(p \land \sim q) \land (\sim p \land q)$
= $(p \land \sim p) \land (\sim q \land q)$
= $f \land f = f$
(By using associative laws and commutative laws).
$∴ (p \land \sim q) \land (\sim p \land q)$ is a contradiction
$∴$ a contradiction holds

Q. (p∧∼q)∧(∼p∧q) is