Question #212772

Show that (p → r) ∨ (q → r) and (p ∧ q) → r are logically equivalent. 


Expert's answer

(p→r)∨(q→r)(p → r) ∨ (q → r)

\equiv (\neg p\lor r) \lor (\neg q\lor r) \ \ \ \ \ \ \ (Implication)

\equiv \neg p\lor r \lor \neg q\lor r \ \ \ \ \ \ \

\equiv (\neg p \lor \neg q)\lor r \ \ \ \ \ \ \ (Distribution)

≡¬(p∧q)∨r\equiv \neg ( p\land q)\lor r (De Morgan's Law)

≡(p∧q)→r\equiv ( p\land q) → r (Implication)

Hence proved ,

(p→r)∨(q→r)(p → r) ∨ (q → r) and (p∧q)→r( p\land q) → r are logically equivalent.





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