# logic and proof

• Jun 23rd 2007, 11:13 AM
TheRekz
logic and proof
What is wrong with this argument? Let S(x,y) b " x is shorter than y". Given the premise $\displaystyle \left( {\exists x} \right)$ S(s,Max), it follows that S(Max,Max). Then by existential generalization it follows that $\displaystyle \left( {\exists x} \right)$S(x,x), so that someone is shorter than himself.
• Jun 23rd 2007, 11:26 AM
topsquark
Quote:

Originally Posted by TheRekz
What is wrong with this argument? Let S(x,y) b " x is shorter than y". Given the premise $\displaystyle \left( {\exists x} \right)$ S(s,Max), it follows that S(Max,Max). Then by existential generalization it follows that $\displaystyle \left( {\exists x} \right)$S(x,x), so that someone is shorter than himself.

It only follows that S(x, Max) if there is such a y such that x < y for all s. This value of y need not exist.

As an example, let the universe of discourse be the set $\displaystyle \{ 1 - n^{-1}| n \in \mathbb{Z}^+ \}$. No Max exists such that S(x, Max) for all x.

-Dan
• Jun 23rd 2007, 11:35 AM
TheRekz
I don't really get your explanation, could you try to explain it a bit more or maybe more examples. Sorry! :(

I don't get the example that you give me, but I understand that in order of S(x,y) to be valid, x has to be less than y and here x = y, so it can't happen right? Am I explaining it the right way? I also don't understand by the domain that you give me. Is that for all x and y?

Is this a type of fallacy?
• Jun 23rd 2007, 11:44 AM
CaptainBlack
Quote:

Originally Posted by TheRekz
What is wrong with this argument? Let S(x,y) b " x is shorter than y". Given the premise $\displaystyle \left( {\exists x} \right)$ S(s,Max), it follows that S(Max,Max). Then by existential generalization it follows that $\displaystyle \left( {\exists x} \right)$S(x,x), so that someone is shorter than himself.

RonL
• Jun 23rd 2007, 12:09 PM
Plato
Quote:

Originally Posted by TheRekz
Is this a type of fallacy?

Yes the fallacy occurred in your first use of existential instantiation.
The basic rule for EI is: $\displaystyle \frac{{\left( {\exists x} \right)\phi (x)}}{{\phi (v)}}$ where v is an individual constant having no prior occurrence in the context.
• Jun 23rd 2007, 12:17 PM
TheRekz
so what type of fallacy is this? it it the affirming the conclusion?
• Jun 23rd 2007, 12:30 PM
Plato
Quote:

Originally Posted by TheRekz
so what type of fallacy is this? it it the affirming the conclusion?

I have no idea what name your instructor/textbook would give to that fallacy. It is just a violation of the instantiation rules.