# [FOM] Re 196:Quantifier complexity in set theory

Kurt Maes maes.kurt at pandora.be
Mon Nov 10 10:23:36 EST 2003

```I think I have found and proven a 5-quantifier expression in "epsilon,="-
notation that is equivalent to the axiom of choice in ZF-set-theory.
Currently my proof is still somewhat in a draft form, and I only have it
.sxw-format ("open-office writer"-document).

The theorem in this document proofs the equivalence of the following 5
sentences:
1) If x is a set of non-empty sets which are pairwise-disjoint, than x has
a choice-set (I.e. the intersection of y with any non-empty element of x is
a singleton.)
2) Same as one, but we do not demand the elements of x to be non-empty.
3) Same as two, but instead of stating the existence of a choice-set, we
sate the existence of a choice-set which is not an element of x itself.
4) for every set x at least one of the following two conditions holds:
a) there exists an element of x which does not contain any element not
already contained in another element of x.
b) x has a choice-set which is not an element of x itself.
5) forall x exists y forall z exists a forall u [y in x and A(x,y,z,a)] or
[y not in x and B(x,y,z,a,u)].
Here A(x,y,z,a) stands for z in y implies (a in x and a not equal y and
z in a)
and B(x,y,z,a,u) stands for z in x implies (u in z implies a in z and a
in y) and (a in z and a in y implies (u in z and u in y implies u=a))

The equivalence of the first four sentences uses ZF.
The equivalence of 4(when rewriten in "epsilon,="-notation) and 5 uses only
logic and nothing of ZF.
Since 1 is known form of the axiom of choice in ZF, 5 is a 5-quantifier
"epsilon,="-expression that is ZF-equivalent with the axiom of choice.

PS.:
I'm currently looking for a Ph.d-stundent position and I have no experience
in publishing work of my own. So, if anyone thinks this result is worth