# Thread: Prove/disprove using logical using logical arguments

1. ## Prove/disprove using logical using logical arguments

Prove or disprove the following using a logical argument.
The universe of discourse for x,y,z is integers.

$\displaystyle \forall [p(x) \lor q(x)] \Leftarrow \forall p(x) \lor \forall q(x)$

2. Originally Posted by aaronrj Prove or disprove the following using a logical argument.
The universe of discourse for x,y,z is integers.

$\displaystyle \forall [p(x) \lor q(x)] \Leftarrow \forall p(x) \lor \forall q(x)$
I don't think what you wrote makes sense. plus writing the arrow backwards looks odd.

do you mean $\displaystyle (\forall x)p(x) \vee (\forall x)q(x) \implies (\forall x)[p(x) \vee q(x)]$ ?

3. Originally Posted by Jhevon I don't think what you wrote makes sense. plus writing the arrow backwards looks odd.

do you mean $\displaystyle (\forall x)p(x) \vee (\forall x)q(x) \implies (\forall x)[p(x) \vee q(x)]$ ?
I looked at all his posts. I don't think he wants help. He seems to be making fun of people by posting rubbish on purpose.

4. Originally Posted by Jhevon 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! 5. Originally Posted by inferno 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.

6. Originally Posted by inferno 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.

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