p - q - p - q isA- A tautologyB- Both a tautology and a contradictionC- A contradictionD- Neither a tautology nor a contradiction

The correct option is C A contradiction(p ∧∼ q)∧ (∼ p ∧ q)=(p ∧∼ p)∧ (∼ q ∧ q) =f ∧ f=f (By using associative laws and commutative laws) ∴ (p ∧∼ q)∧ (∼ p ∧ q) ...

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Byju's Answer

Standard XII

Mathematics

Fallacy

p ∧∼ q ∧∼ p ∧...

Question

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

A tautology

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A contradiction

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Both a tautology and a contradiction

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Neither a tautology nor a contradiction

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Solution

The correct option is B 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.

Suggest Corrections

(p∧∼q)∧(∼p∧q) is

The correct option is B A contradiction(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.

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

The correct option is B 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.