Classical mathematics is regarded as being formalized by ZFC. ZFC starts with a base of first order logic with identity. Whether called 'first order logic with identity', or 'first order logic with equality', or 'identity theory', the usual axioms, whether named as axioms for 'identity' or axioms for 'equality' are as I mentioned. For example, Enderton's 'A Mathematical Introduction To Logic'.
One of the axioms of identity theory is the law of identity formalized: Ax x=x. So the law of identity pertains to ZFC and, if ZFC is consistent, then ZFC does not contradict identity theory.
Ask anyone who studies set theory, whether ZFC is a first order theory with identity. I can't help that the Google doesn't help to find this.
The Wikipedia article you mentioned is not well written. (1) It doesn't give an axiomatization and (2) It doesn't mention that we can have other symbols in the signature and that by a schema we can generalize beyond a signature with only '='.
But, yes, of course we can use the word 'equality' or the word 'identity', as I said so many times. Perhaps 'equality' is used more often, but the exact formal theory is the same, and is axiomatized as I mentioned.
You haven't convinced me that set equality has anything to do with the law of identity. — fishfry
I gave you fulsome explication. Said yet another way:
The axiom of extensionality gives a sufficient condition for equality. But it doesn't give a necessary condition. So it is not a mathematical definition. A formal mathematical definition for a binary predicate R is of the form:
x R y if and only if (something here about x and y)
A definition of '=' would be of the form:
x = y if and only if (something about x and y)
But the axiom of extensionality is usually given:
If, for all z, z is a member of x if and only if z is a member of y, then x= y.
That is of the form:
If (something about x and y) then x = y. And that is only a sufficient condition, not a biconditional.
But I also mentioned that we could have this variation:
x = y if and only if (for all z, z is a member of x if and only if z is a member of y; and for all z, x is a member of z if and only if y is a member of z).
And that would be a definition, since it gives both a sufficient and necessary condition for x = y.
However, I'll say it yet another way:
Ordinarily, set theory is written with a signature of '=' and 'e', where '=' is logical and 'e' is non-logical, and we have the axioms for first order logic with identity (aka 'idenity theory' or 'first order logic with equality'). Then we add the axiom of extensionality. And then we get (1) as a theorem.
No matter which approach we take, we end up with the same theorems.
you claimed that set equality is "identity," — fishfry
For the third time, I did not say that. And I again told you what I do say. Please stop saying that I said something that I did not say.
I said that set theory adheres to the law of identity but that set theory, with its axiom of extensionality, adds an additional sufficient condition for identity.
I don't know why you don't grasp this:
Ax x= x is an axiom of first order logic with identity. And set theory is a theory that subsumes first order logic with identity. So Ax x=x is also an axiom incorporated into set theory. But also set theory adds the axiom of extensionality.
You seem to be saying that. — fishfry
You are egregiously glossing over what I exactly say. So you form incorrect "seems".
If, for whatever reason, there is a point of my terminology that requires definition for you, then, time permitting, I would supply the definition. Or if there is an argument you can't see to be logical, then, time permitting, I would explain it in even more detail if there even is more detail that could be reasonably provided. However, that can lead to a long chain of definitions back to primitive notions, so it would be better to start at the beginning such as in some chosen textbook. But that does not justify you claiming that I said things that I did not say.
But identity theory, merely syntactically, can't require that '=' be interpreted as standing for the identity relation on the universe as opposed to standing for some other equivalence relation on the universe.
— TonesInDeepFreeze
We're not having the same conversation now. I couldn't even parse that. — fishfry
It parses perfectly even if it seems difficult when one is not familiar with the basic mathematical logic in which symbols are interpreted with models. I'll put it this way:
'=' is a 2-place predicate symbol.
A model (an interpretation) for a language assigns a 2-place relation on the universe to a 2-place relation symbol. In other words, that is an assignment of the meaning, per the model, of the 2-place relation symbol. We call that 'the interpretation of the symbol'. That is semantical.
In general, relation symbols are interpreted differently with different models. But in the special case of '=', we stipulate that, with all models, '=' is stands for the identity relation on the universe for the model.
So my point was that from the mere syntactical presence of '=' in a formula, we can't ensure that '=' stands for the identity relation, and we have to turn to semantics (models) for that.
There is no model of set theory in which extensionality is false. — fishfry
Correct. There are no models of the axiom of extensionality in which the axiom of extensionality is false. But there are models in which the axiom of extensionality is false; they are not models of the axiom of extensionality. Take this is steps:
A model M is for a language.
Theories are written in languages.
So if a theory T is written in a language L, then a model for M for L is model for the language of T.
Given a model M, some sentences in the language L are true in M and some sentences in the language L are false in M.
A theory is a set of sentences closed under provability.
If every sentence in theory T is true in a given model M, then we say "M is a model of T".
So, notice that that there is a difference in meaning between "M is a model for the language L" and "M is a model of the theory T".
That is a crucial thing to understand and keep in mind.
Now, back to the axiom of extensionality.
Let T be any theory that is axiomatized with a set of axioms that includes the axiom of extensionality. If M is a model of T, then the axiom of extensionality is true in M.
But there are models for the language for set theory that are not models of set theory. For example
Let M have universe U = {0 1} and let 'e' be interpreted as the empty relation.
M is a model for the language for set theory, and M is not a model of set theory.
Again, to stress:
There is a difference between a language and a theory writtten in that language.
Let L be a language, and T be a theory written in L, and M be a model for L. It is not entailed that M is a model of T.
/
Urelements. Even though search engines are often deficient, I bet that you can find articles about urelements.
Df. x is an urelement <-> (~Ey yex & ~ x = 0) ('0' here standing for 'the empty sety')
A theorem of Z:
~Ex x is an urelement.
But we could have other axioms where
~Ex x is an urlement
is not a theorem.
And we could have axioms where
Ex x is an urlement
is a theorem.
But a theory that has the theorem:
Exy (x is an urelement & y is an urelement & ~ x = y)
is obviously inconsistent with the axiom of extensionality.
But we could do this:
Axiom: Axy((~ x is an urlement & ~ y is an urelement & Az(z e x <-> z e y)) -> x = y)
In sum: Set theory adopts identity theory and the standard semantics for identity theory, and also the axiom of extensionality. With that, we get the theorem:
Axy(x = y <-> Az((z e x <-> z e y) & (x e z <-> y e z)))
And semantically we get that '=' stands for the identity relation.
— TonesInDeepFreeze
Can't possibly be. — fishfry
It be's. Just as I've explained again. And I'll explain yet another way:
Certain proofs, including in set theory, use the identity axioms I mentioned. Set theory also has the axiom of extensionality, which allows for even more proofs. And we have this theorem of set theory:
Axy(x = y <-> Az((z e x <-> z e y) & (x e z <-> y e z)))
That's all purely syntactical.
Meanwhile, we interpret '=' to stand for the identity relation on the universe of any model for the language for set theory. And that is semantical.
Models of what? — fishfry
Models for the language of set theory. Some of them are models of set theory.
Always keep in mind the distinction between "model M for a language L" and "model M of a theory T" (or, mentioning the language also, "model M of a theory T written in a language L").
How do you know whether having the same members is sufficient for identity unless you already have the axiom of extensionality? In which case you're doing set theory, not identity theory. — fishfry
You're very mixed up thinking that you're somehow disagreeing with me on those points.
Indeed, I am saying that the axiom of extensionality is what makes Az(z e x <-> z e y) sufficient for x = y.
And, indeed, identity theory is not set theory. Rather, identity theory is a sub-theory of set theory:
Every theorem of identity theory is a theorem of set theory. But not every theorem of set theory is a theorem of identity theory: Right off the bat, the axiom of extensionality is not a theorem of identity theory.
And again, we needn't quibble that "pure" identity theory does not have 'e' in its signature. I'm talking about the axioms of identity theory written in a signature that includes 'e'.