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 \}.

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