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The two powerhouse schemata of set theory are Replacement and Collection:

Replacement. For every definable function $f$ and every set $x$, $f"x$ is a set.

Collection. For every definable relation $R$ and every set $x$, there is a set $y$ such that for every $u\in x$ there is $v\in y$ such that $u\mathrel{R}v$.

Easily, Collection implies Replacement, and assuming $\sf ZF$ we can prove the converse as well. If we omit the Power Set axiom, then the reverse implication no longer holds, and Collection is a strictly stronger schema than Replacement.

But since $\sf ZF$ without Power Set is strictly weaker, consistency-wise, than $\sf ZF$ itself, it raises the following question:

Does $\sf\operatorname{Con}(ZF(C)-)\to\operatorname{Con}(ZF(C)^-)$?

(Here $\sf ZF-$ is $\sf ZF$ without Power Set, but with Replacement, and $\sf ZF^-$ is the same theory with Replacement replaced by Collection. The C stands for the Well-Ordering Theorem.)

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  • $\begingroup$ You've defined ZF- twice, and ZF^- not at all. $\endgroup$
    – David Roberts ♦
    Commented Jan 4, 2021 at 20:59
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    $\begingroup$ I don't know what you're talking about. $\endgroup$
    – Asaf Karagila ♦
    Commented Jan 4, 2021 at 21:14
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    $\begingroup$ Well, before the edit, anyway :-) $\endgroup$
    – David Roberts ♦
    Commented Jan 4, 2021 at 22:34

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The theories are equiconsistent and have the same strength as second order arithmetic $\text{Z}_2$. Since we have an $L$-definable well-ordering of the constructible universe $L$, replacement implies collection and ZFC\P in $L$.

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  • $\begingroup$ Do we also have the consistency of $\mathsf{ZFC^-}$ with the reflection principle (in the form $\phi(a)\to \exists M\ni a \phi^M(a)$) since $L$ satisfies it? $\endgroup$
    – Hanul Jeon
    Commented Jan 5, 2021 at 5:40
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    $\begingroup$ @HanulJeon Yes, we do. $\endgroup$
    – Dmytro Taranovsky
    Commented Jan 5, 2021 at 17:39

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