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I would like to know if $\forall xx=x$ is an axiom in axiomatic set theory like in other first order languages, or a theorem? If it is a theorem, how to prove it?

Update: In the first order language materials I read,the equality is one of the logical symbols, and $\forall xx=x$ is given as one of the logical axioms. In the ZFC materials I read, the axiom of extensionality is given as $\forall x\forall y(x=y\leftrightarrow\forall z(z\in x\leftrightarrow z\in y))$.

It seems I can have a proof from $\forall x\forall y(x=y\leftrightarrow\forall z(z\in x\leftrightarrow z\in y))$ to $\forall xx=x$ as follows(correct me if I am wrong):

$\forall x\forall y(x=y\leftrightarrow\forall z(z\in x\leftrightarrow z\in y))$

$\forall x\forall x(x=x\leftrightarrow\forall z(z\in x\leftrightarrow z\in x))$

$\forall x\forall x(x=x\leftrightarrow\forall z(T))$

$\forall x\forall x(x=x\leftrightarrow T)$

$\forall x\forall x(x=x)$

$\forall x(x=x)$

If the proof is valid, the logical axiom $\forall x(x=x)$ seems not necessary.

William
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    If the underlying logic is FOL with equality it is a logical axiom. – Mauro ALLEGRANZA Oct 10 '22 at 10:14
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    See also this post as well as this one – Mauro ALLEGRANZA Oct 11 '22 at 07:05
  • @MauroALLEGRANZA Thank you for providing me with the links. I'm still confused about the equality in the logical axiom and the ZFC axiom. Please see the updated question. – William Oct 14 '22 at 14:21
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    Exactly, the FOL with equality assume that equality is an "universal" relation that applies to every sort of objects and also to set. In this case set theory use it without further specifications. – Mauro ALLEGRANZA Oct 14 '22 at 15:07
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    @William It’s true that the axiom of extensionality, stated in that way, implies reflexivity of equality. But I’m not sure that makes the logical axiom “unnecessary”… surely we want to use first order logic with equality for things other than set theory – spaceisdarkgreen Oct 14 '22 at 15:09
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    If instead the logic is without equality, set theory efines it specifically for sets only, with the formula above. In thi case we have to prove that the above definition plus the other set axioms imply the usual equality properties: reflex, symm, trans and substitution. – Mauro ALLEGRANZA Oct 14 '22 at 15:09
  • @MauroALLEGRANZA By "In this case set theory use it without further specifications", do you mean if we use FOL with equality(together with the logical axioms for equality such as reflex, symm, trans and substitution), we do not need the axiom of extensionality any more? – William Oct 15 '22 at 09:22
  • @MauroALLEGRANZA I'm not clear about the difference between FOL with equality(i.e., = defined as a logical symbol) and FOL without equality(i.e., = defined as a non-logical relation symbol).The ZFC materials I read do not say if they are using FOL with equality or FOL without equality, how do you tell the = in the axiom of extensionality is the logical symbol or the relation symbol? And if the set theory does use FOL without equality, can we dispense with the logical axioms about equality safely? – William Oct 15 '22 at 09:23
  • @spaceisdarkgreen Yes, if we apply FOL with equality to other things, we may need the logical axioms about equality. But if we are just talking about ZFC, maybe we can dispense with the logical axioms about equality safely. – William Oct 15 '22 at 09:30
  • @William in first order logic without equality there are no logical axioms about equality in the first place. – spaceisdarkgreen Oct 15 '22 at 18:24

2 Answers2

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It can go either way. By wikipedia, "The converse of this axiom (Axiom of extensionality) follows from the substitution property of equality. If the background logic does not include equality $x=y$ may be defined as an abbreviation for the following formula $\forall z (z\in x\Leftrightarrow z\in y)\wedge\forall w(x\in w\Leftrightarrow y\in w)$". In other words, it doesn't hurt $\in$ to replace $x$ with $y$ whenever $x=y$ (either as the element or the bigger set).

In the latter way, it can be proved trivially since both $z\in x\Leftrightarrow z\in x$ and $x\in w\Leftrightarrow x\in w$ are trivially true, thus $x=x$ follows.

Just a user
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  • By "the background logic does not include equality", do you mean the first order language describing the set theory does not include the equality symbol as one of the logical symbols but include the equality symbol as a relation symbol in the non-logical symbols, or the first order language describing the set theory does not include the equality symbol in either its logical symbols or non-logical symbols, or the the equality symbol is not in the semantic language for the first order language describing the set theory? Please see my updated question. – William Oct 14 '22 at 14:30
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    Again, it can go either way that won't change the theory. $=$ can be introduced as a relation symbol, and $x=y\Leftrightarrow \forall z(z\in x \cdots$ as a new axiom. Or we just don't include $=$ in our language, but use the formula to define it, just like $x\in\mathbb R$ or $y=x+1$ is not in ZFC but has to be defined. – Just a user Oct 15 '22 at 00:35
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Since it is a logical theorem, it can be proven using no assumptions whatsover. Hence, there is no need to include it in any set of axioms.

How to prove it? That depends on the specific rules of the proof system you are working with. Here is a formal proof in Fitch:

enter image description here

Bram28
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