Negation of the statement p - r - q - r is

Negation of (p ∨ r) ⇒ nbsp; (q ∨ r) is nbsp; ∼((p ∨ r) ⇒ nbsp; (q ∨ r)) = ∼( ∼(p r)(q r)) =(p ∨ r) ∧∼ nbsp; (q ∨ r) Shaded region is equivalent to n ...

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Byju's Answer

Standard XI

Mathematics

Negation

Negation of t...

Question

Negation of the statement $(p \lor r) \Rightarrow (q \lor r)$ is

\sim p \land q \land r

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p \land \sim q \land \sim r

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\sim p \land q \land \sim r

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p \land q \land r

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Solution

The correct option is B $p \land \sim q \land \sim r$
Negation of $(p \lor r) \Rightarrow (q \lor r)$ is
$\sim ((p \lor r) \Rightarrow (q \lor r))$
$= \sim (\sim (p \lor r) \lor (q \lor r))$
$= (p \lor r) \land \sim (q \lor r)$

Shaded region is equivalent to
$p \land \sim q \land \sim r$

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Negation of the statement (p∨r)⇒(q∨r) is

The correct option is B p∧∼q∧∼rNegation of (p∨r)⇒(q∨r) is ∼((p∨r)⇒(q∨r)) = ∼(∼(p∨r)∨(q∨r)) =(p∨r)∧∼(q∨r) Shaded region is equivalent to p∧∼q∧∼r

Negation of the statement $(p \lor r) \Rightarrow (q \lor r)$ is

The correct option is B $p \land \sim q \land \sim r$
Negation of $(p \lor r) \Rightarrow (q \lor r)$ is
$\sim ((p \lor r) \Rightarrow (q \lor r))$
$= \sim (\sim (p \lor r) \lor (q \lor r))$
$= (p \lor r) \land \sim (q \lor r)$

Shaded region is equivalent to
$p \land \sim q \land \sim r$