# Prove/disprove using logical using logical arguments

• February 22nd 2010, 06:16 PM
aaronrj
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.

$\forall [p(x) \lor q(x)] \Leftarrow \forall p(x) \lor \forall q(x)$
• February 22nd 2010, 06:35 PM
Jhevon
Quote:

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

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

I don't think what you wrote makes sense. plus writing the arrow backwards looks odd.

do you mean $(\forall x)p(x) \vee (\forall x)q(x) \implies (\forall x)[p(x) \vee q(x)]$ ?
• February 23rd 2010, 06:43 AM
novice
Quote:

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

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

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.
• February 23rd 2010, 06:38 PM
inferno
Quote:

Originally Posted by Jhevon
do you mean $(\forall x)p(x) \vee (\forall x)q(x) \implies (\forall x)[p(x) \vee q(x)]$ ?

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! (Tongueout)
• February 23rd 2010, 07:18 PM
novice
Quote:

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! (Tongueout)

Your friend is lacking the courtesy in acknowledging Jhevon.
• February 24th 2010, 06:29 AM
PiperAlpha167
Quote:

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! (Tongueout)

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.