An equivalence relation on filters

Question

Let $F$ be a filter. We say $ X \sim_{f} Y $ iff $X \leftrightarrow Y$ $\in$ $F$. I am able to prove reflexivity and associativity of the relation but not the transitivity. Need help with that. Use the definition of double implication with join and meet operations.

Answer 1

In logic, we have $X\to Y\ =(\lnot X)\lor Y$, so the corresponding set operation is $X^\complement\cup Y\ \ \left( = (X\setminus Y)^\complement\ \right)$.

Thus, for sets $X\leftrightarrow Y$ would mean $(X^\complement\cup Y)\,\cap\,(Y^\complement \cup X)\ =\ (X^\complement\cap Y^\complement)\,\cup\,(X\cap Y)\ =$
$=\ (X\cup Y)^\complement\,\cup\,(X\cap Y)$.

For elements, we can say that $s\in X\leftrightarrow Y$ iff either $s$ is in both sets, or is in neither of them.

Hint: Prove that $(X\leftrightarrow Y)\ \cap\ (Y\leftrightarrow Z)\ \subseteq\ (X\leftrightarrow Z)$.

Update: In an abstract Boolean algebra, you can also prove this. Perhaps easier to prove first $(X\to Y)\land (Y\to Z)\,\le\, (X\to Z)\ $ (using the definition $A\to B:=\lnot A\lor B$), then we can combine to get $$(X\leftrightarrow Y)\,\land\, (Y\leftrightarrow Z) = \big((X\to Y)\land (Y\to X)\big) \ \land\ \big((Y\to Z)\land (Z\to Y)\big) \ =\\ =\ (X\to Y)\land (Y\to Z)\ \land\ (Z\to Y)\land (Y\to X)\ \le\ (X\to Z)\,\land\,(Z\to X)\,.$$