The correct option is B A contradicition #160; p q #160; ∼ p #160; p ∧ q #160; ∼ p ∧ q #160; #160; (p ∧ q) ∧ (∼ p ∧ q) ...

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Fallacy

p ∧ q ∧ ∼ p ∧...

Question

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

A tautology

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A contradicition

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Tautology and contradiction

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Neither a tautology nor a contradiction

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Solution

The correct option is B A contradicition
$p$ $q$ $\sim p$ $p \land q$ $\sim p \land q$ $(p \land q) \land (\sim p \land q)$
T T F T F F
T F F F F F
F T T F T F
F F T F F F
All the values in the last column are False, hence the given statement is a contradiction.

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