OpenStudy (anonymous):

Prove [(p∧q) → r] → [(p→r)⋁(q→r)] is a tautology without using a truth table.

OpenStudy (paxpolaris):

what other method do you use other than truth table?

OpenStudy (anonymous):

I am not sure. I think we are supposed to use logical equivalences.. Demorgan laws, implication law, etc

OpenStudy (anonymous):

thats why I am asking for help on here

OpenStudy (anonymous):

hello?

OpenStudy (paxpolaris):

left side:\[=\neg r \to\neg ( p \wedge q)\]\[=\neg r \to (\neg p \vee \neg q)\] right side:\[= \left( \neg r \to \neg p \right) \vee \left( \neg r \to \neg q \right)\]

OpenStudy (paxpolaris):

just using implication and deMorgan