Sum
Without using truth table, show that
(p ∨ q) → r ≡ (p → r) ∧ (q → r)
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Solution
L.H.S.
≡ (p ∨ q) → r
≡ ~ (p ∨ q) ∨ r ....[p → q → ~ p ∨ q]
≡ (~ p ∧ ~ q) ∨ r ....[De Morgan’s law]
≡ (~ p ∨ r) ∧ (~ q ∨ r) .....[Distributive law]
≡ (p → r) ∧ (q → r) .....[p → q → ~ p ∨ q]
= R.H.S.
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