Does $\sf ZFA + LIPS$ prove $\sf AC$?

Where $\sf LIPS$ is phrased as: for every infinite set $X$ the set of intervening cardinals between $|X|$ and $|\mathcal P(X)|$ is well-ordered.

In $\sf ZF$, I think $\sf AC$ follows because we'll have a well ordered class of cardinals. But, I suspect this might not the case for $\sf ZFA$?!

I posed this question originally in terms of $\sf GCH$ itself, but this turned to prove $\sf AC$ (Specker ; pp: 419,420). What I had in my mind is to start with a set $A$ of urelements such that $|A|=\aleph_1$, then construct $M=L_\kappa(A) \models \sf ZF$ for some countable $\kappa$, this way $M$ would see $A$ incomparable to any of its pure sets, the reason is because there are no injections available from $A$ to any of the pure sets in $M$ because all of those are externally countable and $A$ is not, on the other hand due to $M$ satisfying Replacement, then Hartogs are definable in $M$ and so any external injection from $\aleph(A)$ to $A$ cannot be an element of $M$. So $M$ sees $A$ as non-well-orderable. If $\sf GCH$ holds, then we'll have $|A| < |A|+ \aleph(A) \leq 2^{|A|}$ leading to $\aleph(A)= 2^{|A|}$, and so well ordering $A$ which cannot be. So we must have $|A| < |A|+ \aleph(A) < 2^{|A|}$. But, given the construction of $M$ then per the above argument all intervening cardinals between $|X|$ and $|\mathcal P(X)|$ are of the form $|A| + \aleph_\alpha$ for $\aleph_\alpha \geq \aleph(A)$, and those are well-orderable! If $M$ sees multiple incomparable subsets of $A$, then the set of intervening cardinals may not be well-orderable. But is this inevitable? I mean can we have a model in which $M$ shuns existence of such incomparable subsets of $A$?

More generally, per specifications of $M$ given above:

Can we have $M \models \sf LIPS$?

Does $\sf ZFA + WOIPS$ prove $\sf AC$?

Where $\sf WOIPS$ is phrased as: for every infinite set $X$ the set of intervening cardinals between $|X|$ and $|\mathcal P(X)|$ is well-ordered.

In $\sf ZF$, I think $\sf AC$ follows because we'll have a well ordered class of cardinals. But, I suspect this might not the case for $\sf ZFA$?!

I posed this question originally in terms of $\sf GCH$ itself, but this turned to prove $\sf AC$ (Specker ; pp: 419,420). What I had in my mind is to start with a set $A$ of urelements such that $|A|=\aleph_1$, then construct $M=L_\kappa(A) \models \sf ZF$ for some countable $\kappa$, this way $M$ would see $A$ incomparable to any of its pure sets, the reason is because there are no injections available from $A$ to any of the pure sets in $M$ because all of those are externally countable and $A$ is not, on the other hand due to $M$ satisfying Replacement, then Hartogs are definable in $M$ and so any external injection from $\aleph(A)$ to $A$ cannot be an element of $M$. So $M$ sees $A$ as non-well-orderable. If $\sf GCH$ holds, then we'll have $|A| < |A|+ \aleph(A) \leq 2^{|A|}$ leading to $\aleph(A)= 2^{|A|}$, and so well ordering $A$ which cannot be. So we must have $|A| < |A|+ \aleph(A) < 2^{|A|}$. But, given the construction of $M$ then per the above argument all intervening cardinals between $|X|$ and $|\mathcal P(X)|$ are of the form $|A| + \aleph_\alpha$ for $\aleph_\alpha \geq \aleph(A)$, and those are well-orderable! If $M$ sees multiple incomparable subsets of $A$, then the set of intervening cardinals may not be well-orderable. But is this inevitable? I mean can we have a model in which $M$ shuns existence of such incomparable subsets of $A$?

More generally, per specifications of $M$ given above:

Can we have $M \models \sf WOIPS$?

Does $\sf ZFA + GCH^{wo}$ prove $\sf AC$?

Where $\sf GCH^{wo}$ is phrased as: for every infinite set $X$ the set of intervening cardinals between $|X|$ and $|\mathcal P(X)|$ is well-ordered.

In $\sf ZF$, I think $\sf AC$ follows because we'll have a well ordered class of cardinals. But, I suspect this might not the case for $\sf ZFA$?!

I posed this question originally in terms of $\sf GCH$ itself, but this turned to prove $\sf AC$ (Specker ; pp: 419,420). What I had in my mind is to start with a set $A$ of urelements such that $|A|=\aleph_1$, then construct $M=L_\kappa(A) \models \sf ZF$ for some countable $\kappa$, this way $M$ would see $A$ incomparable to any of its pure sets, the reason is because there are no injections available from $A$ to any of the pure sets in $M$ because all of those are externally countable and $A$ is not, on the other hand due to $M$ satisfying Replacement, then Hartogs are definable in $M$ and so any external injection from $\aleph(A)$ to $A$ cannot be an element of $M$. So $M$ sees $A$ as non-well-orderable. If $\sf GCH$ holds, then we'll have $|A| < |A|+ \aleph(A) \leq 2^{|A|}$ leading to $\aleph(A)= 2^{|A|}$, and so well ordering $A$ which cannot be. So we must have $|A| < |A|+ \aleph(A) < 2^{|A|}$. But, given the construction of $M$ then per the above argument all intervening cardinals between $|X|$ and $|\mathcal P(X)|$ are of the form $|A| + \aleph_\alpha$ for $\aleph_\alpha \geq \aleph(A)$, and those are well-orderable! If $M$ sees multiple incomparable subsets of $A$, then the set of intervening cardinals may not be well-orderable. But is this inevitable? I mean can we have a model in which $M$ shuns existence of such incomparable subsets of $A$?

More generally, per specifications of $M$ given above:

Can we have $M \models \sf GCH^{wo}$?

Does $\sf ZFA + LIPS$ prove $\sf AC$?

Where $\sf LIPS$ is phrased as: for every infinite set $X$ the set of intervening cardinals between $|X|$ and $|\mathcal P(X)|$ is well-ordered.

In $\sf ZF$, I think $\sf AC$ follows because we'll have a well ordered class of cardinals. But, I suspect this might not the case for $\sf ZFA$?!

I posed this question originally in terms of $\sf GCH$ itself, but this turned to prove $\sf AC$ (Specker ; pp: 419,420). What I had in my mind is to start with a set $A$ of urelements such that $|A|=\aleph_1$, then construct $M=L_\kappa(A) \models \sf ZF$ for some countable $\kappa$, this way $M$ would see $A$ incomparable to any of its pure sets, the reason is because there are no injections available from $A$ to any of the pure sets in $M$ because all of those are externally countable and $A$ is not, on the other hand due to $M$ satisfying Replacement, then Hartogs are definable in $M$ and so any external injection from $\aleph(A)$ to $A$ cannot be an element of $M$. So $M$ sees $A$ as non-well-orderable. If $\sf GCH$ holds, then we'll have $|A| < |A|+ \aleph(A) \leq 2^{|A|}$ leading to $\aleph(A)= 2^{|A|}$, and so well ordering $A$ which cannot be. So we must have $|A| < |A|+ \aleph(A) < 2^{|A|}$. But, given the construction of $M$ then per the above argument all intervening cardinals between $|X|$ and $|\mathcal P(X)|$ are of the form $|A| + \aleph_\alpha$ for $\aleph_\alpha \geq \aleph(A)$, and those are well-orderable! If $M$ sees multiple incomparable subsets of $A$, then the set of intervening cardinals may not be well-orderable. But is this inevitable? I mean can we have a model in which $M$ shuns existence of such incomparable subsets of $A$?

More generally, per specifications of $M$ given above:

Can we have $M \models \sf LIPS$?