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



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> **Replacement.** For every definable function $f$ and every set $x$, $f"x$ is a set.
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> **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 $u\mathrel{R}v$.
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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:



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



Is Collection really stronger than Replacement?
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 $L$.


