Show that the following two statements are equivalent:
(1) $\forall x \neq \emptyset:\forall X \subseteq x: (X \neq \emptyset \implies \exists y \in X:\forall z \in X: z \notin y)$ (I.e. $\in$ is a well-founded relation on $x$).
(2) $\forall x \neq \emptyset: \exists a \in x: x \cap a = \emptyset$
I can show that $(1) \implies (2)$ but I fail to prove $(2) \implies (1)$.
Attempt: Fix $a\in x$ with $x \cap a = \emptyset$.
Let $\emptyset \neq X \subseteq x$. The only thing we can do is fix $y \in X$. Then $y \in x$ and thus $y \notin a$.
If $z \in X$, then how to deduce $z \notin y$? If $z \in y$, I tried to find a contradiction but didn't succeed.
The context of this is ZF.
We assume that for every set $x$ we have $a\in x$ such that $x\cap a=\emptyset$, so let $A$ be a set, and let $X⊆A$.
If $X$ is the empry set, then (1) is vacuously true.
If $X≠\emptyset$, there exists $a_X∈X$ such that $a_X∩X=\emptyset$, so if $b∈X$, then $y∉a_X$, so $X$ has a $∈$-minimal element, i.e. $A$ is well founded by $∈$