Re: free variables in FOL

On Fri, 17 Jun 2005, Ken Quirici wrote:

> Jim Spriggs wrote:
> >
> > Consider this rule
> >
> > from F(x) => G
> > deduce (Ax)F(x) => G
> >

(x)(F(x) -> G)
(x)F(x) -> (x)G
(x)G -> G
(x)F(x) -> G

> > Can you find a counter example?
>
No.

> Universe of discourse:
>
> set of integers I
>
> F(x): x is divisible by 5
>
> G: there are primes divisible by 5
>
> then clearly F(x) => G for x=5, but not for any other x in I.
>
So what? It's not a counteraxample

Since G
P -> G for any P

.

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