Dual of- p - q-p- q- r-

The correct option is A p ∨[∼ q∧(p∨ q)∧∼ r] For dual, ⇒ ' ∨ ' is replaced by ' ∧ ' #160; #160; #160; #160; #160; #160; #16 ...

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Contra-Positive

Dual of: p ...

Question

Dual of: $p \land [\sim q \lor (p \land q) \lor \sim r]$

p \lor [\sim q \land (p \lor q) \land \sim r]

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p \lor [\sim q \lor (p \land q) \lor \sim r]

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p \land [\sim q \lor (p \lor q) \lor \sim r]

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p \land [\sim q \lor (p \land q) \lor r]

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Solution

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The dual of statement $p \lor (q \land r) \equiv (p \lor q) \land (p \lor r)$ is

Factorise: $(p^{2} + q^{2} - r^{2})^{2} - 4 p^{2} q^{2}$

(p - q)^{3} + (q - r)^{3} + (r - p)^{3}

The contrapositive of $(\sim p \land q) \to (q \land \sim r)$ is

Factorise: p^{2} + q^{2} + 2 (p q + q r + r p) .

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