Boolean Prime Ideal Theorem/Proof 2

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Theorem

Let $\struct {S, \le}$ be a Boolean algebra.

Let $I$ be an ideal in $S$.

Let $F$ be a filter on $S$.

Let $I \cap F = \O$.

Then there exists a prime ideal $P$ in $S$ such that:

$I \subseteq P$

and:

$P \cap F = \O$

Proof

We prove that the Boolean Prime Ideal Theorem is equivalent to the Ultrafilter Lemma in ZF.

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