Set theory and identity theory
- From: MoeBlee <jazzmobe@xxxxxxxxxxx>
- Date: Thu, 13 Mar 2008 16:04:12 -0700 (PDT)
There's are some questions in my mind now that I don't know how to
settle. They're rather fundamental, so I feel some urgency to get them
straightened out.
First, let's axiomatize identity theory (for a language with at least
one predicate symbol) as folllows (leaving off leading universal
quantifiers as I'll do in this post):
Your favorite axioms (and/or rules) of pure first order predicate
logic (in the abovementioned language), plus
Axiom: x=x
Axiom schema: If P is an atomic formula and P' is the same as P except
y occurs in zero or more places in P' where x occurs in P, then
x=y -> (P -> P')
is an axiom.
We'll call those 'the axioms of identity theory' (though identity
theory may be axiomatized in other ways too).
Now consider a set theory such as Z.
We can take Z to be a non-conservative extension of identity theory
for the language with just the primitive 'e' (aside from the primitive
'='). So this axiomatization has two primitives - '=' and 'e' and the
axioms of identity theory, plus the axiom of extensionality, plus the
other axioms of Z. Let's call this axiomatization Z1. And ordinarily
along with this we would take the fixed semantics that always
inteprets '=' as standing for the identity relation on the universe of
the model.
Or we can dispense with identity theory and the axiom of
extensionality and have instead the pure logical axioms and/or rules
and the rest of the axioms of Z plus the following axiom:
Axiom: Az(zex <-> yex) -> Az(xez -> yez).
So this axiomatization has only one primitive - 'e'. We'll call this
axiomatization Z2.
Then we conservatively extend the theory axiomatized by Z2 by adding
this definitional axiom (which is actually the ordinary axiom of
extensionality):
Def: x=y <-> Az(zex <-> yex)
Now this axiomatization has two non-logical symbols 'e' and '='. We'll
call this axiomatization Z3.
Now, if we take 'theory' in the sense of 'a set of sentences closed
under entailment', then the consequences of Z1 = the consequences of
Z3. They are the exact same set of sentences. That is, Z1 and Z3
axiomatize the same theory. In particular both have the theorems:
x=y <-> Az(zex <-> zey)
and
x=y <-> Az(xez <-> yez).
Now, I want to be quite literal and take the consequences of Z3 as a
theory in a language with both '=' and 'e' so that a structure for the
language gives an interpretation of both '=' and 'e'. Of course, since
'=' is defined from a previous axiomatization, it would suffice to
interpret just 'e'; but for the purpose of drawing out my concern
here, it makes things more visible to also explicitly interpret '='.
But what about the interpretation of '=' when we've used
axiomatization Z3? In this case, since we are not "coming from"
identity theory, I would think we wouldn't give '=' the special
dispensation of a special semantics. In this case, '=' is a defined
symbol, and it gets interpreted however it may be interpreted. But
then is it the case that for any model of the consequences of Z3, it
turns out ANYWAY that '=' gets interpreted as identity on the universe
of the model? (I was unable to prove it when I sat down to the problem
last night.) What about when the model is standard (i.e., interprets
'e' as the membership relation on the universe)? What about when the
model is non-standard (interpretes 'e' as other than the membership
relation on the universe)?
Additionally, with such theories as the consequences of Z1 and the
consequences of Z3, we get the indiscernibliity of identicals (which
is expressed by the axiom schema of identity theory, which is a
theorem schema of Z3). But, as is famous, it is not possible in first
order to express the identity of indiscernibles (not even in a meta-
language for a fist order theory). However with Z1 and Z3 don't we
"override" that in the sense that we have:
Az(zex <-> zey) -> x=y.
That tells us that to conclude x=y, we don't even have to exhaust ALL
properties expressible by formulas, but rather that just one single
property (expressed by the left side fo the above biconditional). For
that matter, we get the same result from another theorem we have:
Az(xez <-> yez) -> x=y.
So, to recap, my questions are:
(1) Is it the case that for any model of the consequences of Z3, it
turns out ANYWAY that '=' gets interpreted as identity on the universe
of the model? What about when the model is standard (i.e., interprets
'e' as the membership relation on the universe)? What about when the
model is non-standard (interpretes 'e' as other than the membership
relation on the universe)?
(2) Though not precise, is my general take on the identity of
indiscernibles in the right vein regarding this particular situation?
Thanks for any help.
MoeBlee
.
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