Equivalence of Versions of Axiom of Choice/Formulation 1 implies Formulation 3

Theorem
The following formulation of the Axiom of Choice:

Formulation 1
implies the following formulation of the Axiom of Choice:

Proof
Let $\SS$ be the set:


 * $\SS = \set {s: \O \notin s \land \forall t, u \in s: t = u \lor t \cap u = \O}$

Let $c$ be a choice function on $\SS$ and consider the image set $c \sqbrk \SS$:


 * $c \sqbrk \SS = \set {\map c s: \O \notin s \land \forall t, u \in s: t = u \lor t \cap u = \O}$

By the definition of choice function:
 * $\map c s \in s$

By construction of $\SS$, for any $s \in \SS$:
 * $s \cap c \sqbrk \SS = \set {\map c s}$