The proposition p - approx-pq- is equivalent to

Explanation for the correct option:Given proposition, p→ (p∧ q)= (p∧ q)→ p∨ q 0ex0ex=p→ (p ∨ q)0ex0ex=( p)∨ ( p∨ q)0ex0ex=( p)∨ qHence, the correct option is D ...

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Standard IX

Mathematics

Property 5

The propositi...

Question

The proposition $p \to ~ (p \land q)$ is equivalent to

$(~ p) \lor (~ q)$

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$(~ p) \land q$

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$q$

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$(~ p) \lor q$

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Solution

The correct option is D
$(~ p) \lor q$

Explanation for the correct option:
Given proposition, $p \to ~ (p \land q)$
$= ~ (p \land ~ q) \to ~ p \lor q = p \to (p ~ \lor q) = (~ p) \lor (~ p \lor q) = (~ p) \lor q$
Hence, the correct option is D, $(~ p) \lor q$

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Similar questions

Q. The proposition

\sim (p \Leftrightarrow q)

is equivalent to

Q. The proposition

\sim (p \lor \sim q) \lor \sim (p \lor q)

is logically equivalent to :

Q. The proposition

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

is equivalent to.

Q. The negation of the compound proposition

p \land (\sim p \lor q)

is equivalent to

Q. The logically equivalent proposition of

p \Leftrightarrow q

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The proposition p→~(p∧q) is equivalent to

The correct option is D ~p∨qExplanation for the correct option:Given proposition, p→~(p∧q)=~(p∧~q)→~p∨q=p→(p~∨q)=~p∨(~p∨q)=~p∨qHence, the correct option is D, ~p∨q

The proposition $p \to ~ (p \land q)$ is equivalent to

The correct option is D
$(~ p) \lor q$

Explanation for the correct option:
Given proposition, $p \to ~ (p \land q)$
$= ~ (p \land ~ q) \to ~ p \lor q = p \to (p ~ \lor q) = (~ p) \lor (~ p \lor q) = (~ p) \lor q$
Hence, the correct option is D, $(~ p) \lor q$