This question is closely connected to the following paper and to a prior question posted to Mathoverflow titled "Can Cantor's theorem be proved in Paraemter Free Zermelo",


That paper provided an answer to a question about whether $\text{ZFC}^o$ is equivalent to $\text{ZFC}$, where $\text{ZFC}^o$ is the theory axiomatized by axioms of Extensionality, Foundation, Pairing, Union, Power, Infinity and a Parameter free version of Replacement and Separation denoted by $\text{Repl}^o$, $\text{Aus}^o$ respectively; where the axioms of Pairing, Union and Power are written in full, i.e. one can prove existence the sets $\{a,b\}$; $U(a)$; $P(a)$ for any sets a,b using those three axioms and Extensionality without using an instance of schema of specification.

In the Wikipedia, axioms of $\text{ZFC}$ are written in a different manner, see the following link:


The axioms of Pairing, Union, Power and Replacement only asserts existence of a non specific set that contains sets a,b ; elements of elements of a; subsets of a, replacements of elements of a respectively; one needs to use Specification in order to prove existence of the sets $\{a,b\}$; $U(a)$; $P(a)$, $F(a)$

Now if by $\text{eZFC}^o$, denoting $\text{"extreme Parameter Free ZFC"}$, it is meant $\text{ZFC}$ axiomatically presented as how it is mentioned in the Wikipeida but with Specification and Replacement axioms replaced by parameter free versions of them, i.e. by $\text{Repl}^o$ and $\text{Aus}^o$ with these two schemata being formalized exactly as in the above-mentioned article (see: Postscript) for full exposition of $\text {eZFC}^o$. Then apparantly the proof presented in the above article fails, because it clearly depends on specifying pairs, unions, powers and replacment sets without the need to use Specification, and thus in some manner it allows the use of Parameters external to Specification scheme. Along the same line $\text{eZC}^o$ denotes $\text{"extreme Parameter Free Zermelo"}$, and this is $\text{eZFC}^o$ minus Replacment.

Given that, the questions are:

Is $\text{eZFC}^o$ equivalent to $\text{ZFC}$?

Is $\text{eZC}^o$ equivalent to $\text{ZC}$?

If both are false then is Cantor's theorem provable in $\text{ZFC}^o$ or $\text{ZC}^o$?

PS: $\text{eZFC}$ is formulated in the first order language of set theory with the following axioms:

Extensionality: $\forall A,B (\forall X (X \in A \iff X \in B) \iff \forall Y (A \in Y \iff B \in Y))$

Foundation: $\forall A (\exists X \in A \implies \exists B \in A (\not \exists C \in B (C \in A)))$

Specification: if $\phi$ is a formula in which only symbol $X$ occurs free, then

$\forall A \exists B \forall X (X \in B \iff X\in A \wedge \phi)$

is an axiom.

Pairing: $\forall A,B \exists C (A \in C \wedge B\in C)$

Union: $\forall A \exists B \forall X,Y (X \in Y \wedge Y \in A \implies X \in B)$

Power: $\forall A \exists B \forall X (\forall Y \in X (Y \in A) \implies X \in B)$

Replacement: If $\phi(Y,Z)$ is a formula in which only symbols $Y,Z$ occur free, then:

$\forall A \exists B \forall X (\exists Y \in A (\forall Z (\phi(Y,Z) \iff Z=X)) \implies X \in B)$

is an axiom.

Infinity: $\exists X (\varnothing \in X \wedge \forall Y \in X (\{Y\} \in X))$

$\text {+\-}$



1 Answer 1


Since Replacement can be proved from parameter-free Replacement (a theorem of Azriel Levy), and Separation can be proved in full from Replacement, the answer is that if we have parameter-free Replacement, and $\sf ZF$ without Replacement and Separation, then we can prove Replacement, and therefore Separation as well.

Azriel Levy, "Parameters in the comprehension axiom schemas of set theory", in: Leon Henkin et al. (Eds.), Proceedings of the Tarski Symposium, in: Proceedings of Symposia in Pure Mathematics, vol. 25, American Mathematical Society, Providence, 1974, pp. 309–324.

Asking about Zermelo (which I presume means $\sf ZF$ without Replacement, but with Separation) amounts to your previous question about parameter-free separation.

  • $\begingroup$ Does Azriel Levy proof of Replacement from parameter free Replacement uses full pairing, union, power or full replacement? Can you refer me to the proof of that theorem of Azriel Levy? $\endgroup$ Nov 21, 2016 at 5:26
  • 1
    $\begingroup$ I've added a reference, but I don't know where to find the article online. I did ask Azriel once if he has a copy, or even a draft, but he said that he didn't see any, and would let me know if he finds something like that. Chances are slim, though. $\endgroup$
    – Asaf Karagila
    Nov 21, 2016 at 7:02
  • $\begingroup$ See the article I've refered to in my answer. It shows the contrary result. $\endgroup$ Nov 21, 2016 at 8:43
  • $\begingroup$ What the paper I referred to in my answer above is saying comes in conflict with what Philip Schlicht is saying in the paper wwwmath.uni-muenster.de/u/rds/ZFC_without_parameters.pdf, Unless what is meant by pairing, union and power in Levy's proof is the implicational form of those axioms (like the ones used in the postscript of the question of this posting) $\endgroup$ Nov 21, 2016 at 18:20
  • $\begingroup$ the reference you've mentioned is present electronically here: bookstore.ams.org/pspum-25?c=1&e=1, it costs around $74 to electronically purchase it. $\endgroup$ Nov 21, 2016 at 18:44

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