# One more conjecture about provability without axiom of choice

I addition to this conjecture I formulate one more similar conjecture:

Conjecture $a\setminus^{\ast} b = a\#b$ for arbitrary filters $a$ and $b$ on a powerset cannot be proved in ZF (without axiom of choice).

Notation (where $\mathfrak{F}$ is the set of filters on a powerset ordered reverse to set-theoretic inclusion):

• $a\setminus^{\ast} b = \bigcap\{z\in\mathfrak{F} \,|\, a\subseteq b\cup z \}$;
• $a\#b = \bigcup\{z\in\mathfrak{F} \,|\, z\subseteq a\wedge z\cap b=0 \}$.