Question
Without using truth table show that:
$p \leftrightarrow q=(p \wedge q) \vee(\sim p \wedge \sim q)$
$p \leftrightarrow q=(p \wedge q) \vee(\sim p \wedge \sim q)$
L.H.S = p ↔ q
≡ (p → q) ∧ (q → p) ........(Biconditional Law)
≡ (∼ p ∨ q) ∧ (∼ q ∨ p) ........(Conditional Law)
≡ [∼ p ∧ (∼ q ∨ p)] ∨ [q ∧ (∼ q ∨ p)] ....(Distributive Law)
≡ [(∼ p ∧ ∼ q)] ∨ (∼ p ∧ p)] ∨ [(q ∧ ∼ q) ∨ (q ∧ p)] .........(Distributive Law)
≡ [(∼ p ∧ ∼ q) ∨ F] ∨ [F ∨ (q ∧ p)] ........(Complement Law)
≡ (∼ p ∧ ∼ q) ∨ (q ∧ p) .......(Identity Law)
≡ (∼ p ∧ ∼ q) ∨ (p ∧ q) ........(Commutative Law)
≡ (p ∧ q) ∨ (∼ p ∧ ∼ q) ........(Commutative Law)
≡ R.H.S.
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$\text{x}_\text{i}$
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$-5$
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$-4$
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$1$
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$2$
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$\text{p}_\text{i}$
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$\frac{1}{4}$
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$\frac{1}{8}$
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$\frac{1}{2}$
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$\frac{1}{8}$
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