Re: ZFC means?



G. Frege wrote: 
On Fri, 27 Jan 2006 05:45:03 -0600, David C. Ullrich
<ullrich@xxxxxxxxxxxxxxxx> wrote:


And in every treatment I've ever seen ZFC _is_ defined
to be a certain theory in FOL with equality. Or
WITH EQUALITY - I suppose that makes it more clear.


Sure. But that doesn't mean that there is no other way to deal with
identity in set theory (say ZFC).


In ZFC, it is IMPORTANT that you are in first-order logic
WITHOUT equality, because in ZF, you can DEFINE equality:
in ZF, x=y MEANS Az[zex<->zey]


Of course george's usual overgeneralization. :-(

USUALLY we use /FOPL with identity/ for ZFC. But it's POSSIBLE to use
just FOPL (without identity) too, in the latter case we may define
equality.


Actually it's IMPORTANT that ZFC be defined as a theory
in FOL with equality, for something kinda more or less
exactly this reason. 


No. But of course it's rather convenient to do so. (And of course, it
is the standard way, I guess.)


The very first axiom listed is Extensionality, which
states exactly that

AxAy(x=y <-> Az(z e x <-> z e y)).

That's not given as a definition of equality, it's
an _axiom_. A perfectly standard axiom, included in
every treatment of ZFC I've ever seen.


It's possible to introduce the definition (of "=")

	x = y :<-> Az(z e x <-> z e y)

to the same effect.


Now tell me: How do you _state_ that axioms in FOL
without equality?


From the definition we get the THEOREM:

	AxAy(x = y <-> Az(z e x <-> z e y))

- that's just as good as the usual AXIOM.

I seem to be puzzled about something. So if ZFC is formalized
in FOL with identity, we still could define x = y as has been
demonstrated, which could then be used to prove x = x as a special
case of the = definition. But isn't it true that x = x is already
given in FOL with identity, and needs no definition? Thanks for a
clarification on this.

---Nam


F.


--
"I do tend to feel Hughes & Cresswell is a more authoritative
 source than you." (David C. Ullrich)


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