Restricted Tukey's Theorem/Strong Form

Theorem
Let $X$ be a set.

Let $\mathcal A$ be a non-empty set of subsets of $X$.

Let $'$ be a unary operation on $X$.

Let $\mathcal A$ have finite character.

For all $A \in \mathcal A$ and all $x \in X$, let either:
 * $A \cup \left\{ {x}\right\} \in \mathcal A$

or:
 * $A \cup \left\{ {x'}\right\} \in \mathcal A$

Then for each $A \in \mathcal A$ there exists a $C \in \mathcal A$ such that:
 * $A \subseteq C$

and:
 * for all $x \in X$, either $x \in C$ or $x' \in C$.

Proof
Let $A \in \mathcal A$.

Let:
 * $\mathcal B = \left\{ {B: B \subseteq X \text{ and } A \cup B \in \mathcal A}\right\}$

It is to be shown that $\mathcal B$ has finite character:

First suppose that $B \in \mathcal B$ and $F$ is a finite subset of $B$.

Then since $B \in \mathcal B$, $B \subseteq X$ and $A \cup B \in \mathcal A$.

We wish to show that $F \in \mathcal B$.

Since $F \subseteq B \subseteq X$:
 * $F \subseteq X$

It remains to be shown that:
 * $A \cup F \in \mathcal A$.

$A \cup F \subseteq A \cup B$.

Let $G$ be a finite subset of $A \cup F$.

Then $G$ is a finite subset of $A \cup B$.

Since $A \cup B \in \mathcal A$ and $\mathcal A$ has finite character, $G \in \mathcal A$.

Thus every finite subset of $A \cup F$ is in $\mathcal A$.

Since $\mathcal A$ has finite character, $A \cup F \in \mathcal A$.

Thus $F \in \mathcal B$.

Suppose instead that $B \subseteq X$ and every finite subset of $B$ is an element of $\mathcal B$.

We wish to show that $B \in \mathcal B$.

In order to do this, we must show that $A \cup B \in \mathcal A$.

Let $F$ be a finite subset of $A \cup B$.

$\mathcal A$ has finite character, $F \cap A \in \mathcal A$.

Since $F \cap B$ is a finite subset of $B$, $F \cap B \in \mathcal B$ by assumption.

Then by the definition of $\mathcal B$:
 * $\left({F \cap B}\right) \cup A \in \mathcal A$

But $F$ is a finite subset of $\left({F \cap B}\right) \cup A \in \mathcal A$.

Since $\mathcal A$ has finite character, $F \in \mathcal A$.

As this holds for all finite subsets of $A \cup B$ and $\mathcal A$ has finite character:
 * $A \cup B \in \mathcal A$

Let $B \in \mathcal B$ and $x \in X$.

Then:
 * $B \cup A \in \mathcal A$

so either $B \cup A \cup \left\{ {x}\right\}$ or $B \cup A \cup \left\{ {x'}\right\}$ is in $\mathcal A$.

But then $B \cup \left\{ {x}\right\}$ or $B \cup \left\{ {x'}\right\}$ is in $\mathcal B$ by the definition of $\mathcal B$.

Thus $\mathcal B$ satisfies the premises of the weak form of the Restricted Tukey-Teichmüller Theorem.

Thus there is a $B \in \mathcal B$ such that for all $x \in X$, either $x \in B$ or $x' \in B$.

Let $C = A \cup B$.

Then since $B \subseteq C$, if $x \in X$ then either $x \in C$ or $x' \in C$.

But since $B \in \mathcal B$:
 * $C = A \cup B \in \mathcal A$