# Existential Instantiation

• October 11th 2008, 12:45 PM
bclere
Existential Instantiation
I need to prove or disprove the following: Z = integers
If x,y ÎZ, with x=2q and y=2h+1, for some q,hÎZ, then \$uÎZ such that xy=2u+1

It seems like existential instantiation is the only way to do this. I don't know how to show it. Can I simply let h = u since addition and multiplication are closed under Z?

Thanks
• October 11th 2008, 01:12 PM
Jhevon
Quote:

Originally Posted by bclere
I need to prove or disprove the following: Z = integers
If x,y ÎZ, with x=2q and y=2h+1, for some q,hÎZ, then \$uÎZ such that xy=2u+1

It seems like existential instantiation is the only way to do this. I don't know how to show it. Can I simply let h = u since addition and multiplication are closed under Z?

Thanks

Hint: if $x = 2q$ and $y = 2h + 1$ then $xy = 2q(2h + 1) = \cdots$
• October 11th 2008, 01:52 PM
bclere
existential instantiation
= what? That doesn't show proof of anything!
• October 11th 2008, 02:30 PM
Jhevon
Quote:

Originally Posted by bclere
= what? That doesn't show proof of anything!

how do you figure? what i typed is everything you need

the question you are supposed to ask yourself now is: can i write 2q(2h + 1) in the form 2u + 1 using only integers?

if the answer is yes, then do it and hence prove the theorem

if the answer is no, then say why you can't do it, thus disproving the theorem