Re: FO logic without equality

On Thu, 01 Jun 2006 15:53:37 +0200, Jan Burse <janburse@xxxxxxxxxxx>
wrote:

Hi

David C. Ullrich wrote:
Huh? Nobody has disputed the fact that
_that_ sentence has only models of cardinality 1.
in fact it was exactly the same model (posted
before I saw his post - it's the natural example.)

If you're going to quote something someone said
do it properly. The juxtaposition of those two
sentences is incoherent (as opposed to what I
actually wrote.)

Maybe this is of interest to some newbees in
model theory. The "natural example" which gives
an infinite model M1 to the sentence in FOL without
builtin equality but with equality axioms:

forall x (x=c) or written without the infix
shorthand = as
forall x(EQ(x,c))

can be domain(M1) = {0,1,2,..}, M1(c)=0, M(EQ)=N x N.
But there can also be "natural examples" which do not
involve infinity but show that FOL without builtin
equality but with equality axioms is different from FOL
with builtin equality.

Yes, maybe this is of interest to newbees. Or maybe
the only person who didn't realize that FOL with
equality was different from FOL without equality
plus axioms was you. It was _you_ who said

"One can define in a FOL without equality,
the equality. This can be found in
every standard text book about logic.

So it doesn't matter wether = is builtin
or whether you have a relation symbol EQ
with the properties of =."

I've never seen anyone else make any such claim.
Today it turns out it's not so - I guess they're
going to have to revise all those standard texts...

************************

David C. Ullrich
.

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