do you mean $\displaystyle (\forall x)p(x) \vee (\forall x)q(x) \implies (\forall x)[p(x) \vee q(x)]$ ?
Yes Jhevon, that is right, our substitute instructor corrected that homework problem today. I am in the same class as the original poster, so I assure you, he is not just posting rubbish!
Yes Jhevon, that is right, our substitute instructor corrected that homework problem today. I am in the same class as the original poster, so I assure you, he is not just posting rubbish!
Your friend is lacking the courtesy in acknowledging Jhevon.
Yes Jhevon, that is right, our substitute instructor corrected that homework problem today. I am in the same class as the original poster, so I assure you, he is not just posting rubbish!
It's a theorem of first-order predicate logic.
Proof?
First thing to notice is that the antecedent is a disjunction.
Informally, think proof by cases.
Or formally, say in an NDS, think VE.