The correct option is B A contradiction p amp; q amp; ∼ p amp; p⇒ q amp; ∼( p⇒ q ) #160; amp; ∼ p∨ q amp; ∼( p⇒ q ) ⟺ p∨ q T am ...

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Combination

Solve: ∼ p⇒ ...

Question

Solve:
$\sim (p \Rightarrow q) \Leftrightarrow\sim p \lor q$

A tautology

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A contradiction

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Neither a tautology nor a contradiction

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Cannot come to any conclusion

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Solution

The correct option is B A contradiction
$\begin{matrix} p & q & \sim p & p \Rightarrow q & \sim (p \Rightarrow q) & \sim p \lor q & \sim (p \Rightarrow q) ⟺ p \lor q T & T & F & T & F & T & F T & F & F & F & T & F & F F & T & T & T & F & T & F F & F & T & T & F & T & F \end{matrix}$

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Similar questions

(p \land \sim q) \land (\sim p \land q)

(p \land \sim q) \land (\sim p \land q)

(p \land \sim q) \land (\sim p \land q)

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

[(p \Rightarrow q) \land (q \Rightarrow r)] \Rightarrow (p \Rightarrow r)

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