A Axiom of choice: Replacing a strong condition with a weaker condition

[jose diez]

This set-theory theorem is very easy to prove:
(*) if A≈B & C≈D & A∩C=∅ & B∩D=∅ then A∪C≈B∪D
It seems intuitive that if one replaces the strong
A∩C=∅ & B∩D=∅
condition by the weaker
A∩C≈B∩D
the implication
(**) if A≈B & C≈D & A∩C≈B∪∩D then A∪C≈BD
still holds.
(**) does not seem to be much stronger than (*), nevertheless I have been able to prove (**) only using Ax of Choice (ACh). This suggested to me that (**) might be other equivalent to ACh, but I have not found it in the standard lists, nor I have been able to prove that (**) implies ACh.
Does anybody have any clue on this?

 

[jose diez]

Sorry there was a typo in (**)

Corrected:
(**) if A≈B & C≈D & A∩C≈B∩D then A∪C≈B∪D

 

[Office_Shredder]

What do the squiggly equal signs stand for? Equal cardinality?

 

[jose diez]

yes

 

[pbuk]

Does anybody have any clue on this?

I haven't worked through it, but isn't it possible to prove both (*) or (**) by induction without invoking AC? In this case, they would be independent of AC so cannot be equivalent.

 

[nuuskur]

Corrected:
(**) if A≈B & C≈D & A∩C≈B∩D then A∪C≈B∪D

Shooting from hip, I'd say choice is not necessary for this to occur. To prove choice, we would have an arbitrary family of nonempty sets and we must find a choice function. The condition (**) is formulated in only finite terms. I don't see how it provides an angle to tackle with infinite families.

That said, I don't have a counterexample.

 

[Office_Shredder]

If you subtract $A\cap C$ from the left side of everything, and $B\cap D$ on the right side, then you have reduced to the case of no intersection as long as you can show the following result:

If $B\subset A$, $D\subset C$, $D\approx B$, $A\approx C$, then $A-B \approx C- D$