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Consider the following forms of the axiom of choice:

$AC_1:\forall F\neq \emptyset~~~(\emptyset\notin F~\wedge~\forall x,y\in F~~~(x\neq y\rightarrow x\cap y= \emptyset))\rightarrow \exists C~\forall x\in F~~~~|C\cap x|=1$

$AC_2:\forall x~\exists r\subseteq x\times x~~~;\langle x,r\rangle~\text{is a well-ordering}$

It is well-known that within $ZF-AF$ (i.e. $ZF$ with the axiom of foundation deleted) these two forms of the axiom of choice are equivalent.

Also in $ZF-AF-P$ (i.e. $ZF-AF$ with power set axiom deleted) the axiom $AC_2$ implies $AC_1$ (See Lemma $I.12.3$ in page $69$ of Kunen's new set theory book) but it seems in the inverse direction the use of the power set axiom to prove $AC_2$ using $AC_1$ is essential. (See Lemma $I.12.4$ in page $69$ of Kunen's book).

Kunen in page $19$ of his book says: "the proof of that $AC_1$ implies $AC_2$ seems to require the power set axiom." I didn't find any proof of this claim.

Question: Is Kunen's intuition true? Is $AC_2$ strictly stronger than $AC_1$ within $ZF-AF-P$? In the other words, assuming consistency of $ZF$ do we have $ZF-AF-P+AC_1\nvdash AC_2$? Precisely, is the following true?

$$Con(ZF)\Longrightarrow Con(ZF-AF-P+AC_1+\neg AC_2)$$

Remark: We have: $Con(ZF)\Longrightarrow Con(ZF-AF-P-Rep+AC_1+\neg AC_2)$. To see this fact it suffices to consider model $M$ of $ZF-AF-P-Rep+AC_1+\neg AC_2$ as follows:

$M_{0}:=HF\cup\{\omega\}$

$\forall n\in \omega~~~~~M_{n+1}:=M_{n}\cup [M_{n}]^{<\omega}\cup \{\cup x~|~x\in M_{n}\}\cup \{x~|~\exists y\in M_{n}~x\subseteq y\}$

$M:=\bigcup_{n\in \omega}M_{n}$

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