The compound propositionP- q- p - q is

P ∧ q) → (p ∨ q) ⇔∼ (p ∧ q) ∨ (p ∨ q) [∵ (a → b) nbsp; ⇔ (∼ a ∨ b)] ⇔ (∼ p ∨∼ q) ∨ (p ∨ q) [ By Demorgan's law] ⇔ (∼ p ∨ p) ∨ (∼ q ∨ q) [By Commutative ...

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Byju's Answer

Standard XI

Mathematics

Compound Statement

The compound ...

Question

The compound proposition
$(P \land q) \to (p \lor q) i s$

a contradiction

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none of these

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a contingency

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a tautology

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Solution

The correct option is D a tautology
$p \land q) \to (p \lor q) \Leftrightarrow\sim (p \land q) \lor (p \lor q) [∵ (a \to b) \Leftrightarrow (\sim a \lor b)] \Leftrightarrow (\sim p \lor \sim q) \lor (p \lor q) [B y D e m o r g a n^{'} s l a w] \Leftrightarrow (\sim p \lor p) \lor (\sim q \lor q) [B y C o m m u t a t i v e a n d A s s o c i a t i v e l a w s] \Leftrightarrow T \lor T \Leftrightarrow T$

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Similar questions

The negation of the compound statement $p \lor (\sim p \lor q)$ is ___.

Q. The compound statement

(P \lor Q) \land (\sim P) \Rightarrow Q

is equivalent to :

Q. The negation of the compound proposition

p \lor (\sim p \lor q)

Q. If

S^{*} (p, q)

is the dual of the compound statement S (p, q) then

S^{*} (\sim p, \sim q)

is equivalent to

Q. The negation of the compound proposition

p \lor (\sim p \land q)

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The compound proposition (P∧q)→(p∨q) is

The correct option is D a tautologyp∧q)→(p∨q)⇔∼(p∧q)∨(p∨q) [∵(a→b)⇔(∼a∨b)]⇔(∼p∨∼q)∨(p∨q) [By Demorgan′s law]⇔ (∼p∨p)∨(∼q∨q) [By Commutative and Associative laws]⇔ T∨T⇔ T

The compound proposition
$(P \land q) \to (p \lor q) i s$