An equivalence relation on filters 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.
 A: 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
$$(X\leftrightarrow Y)\,\land\, (Y\leftrightarrow Z) = \big((X\to Y)\land (Y\to X)\big) \ \land\ \big((Y\to Z)\land (Z\to Y)\big) \ =\\
=\ (X\to Y)\land (Y\to Z)\ \land\ (Z\to Y)\land (Y\to X)\ \le\ 
(X\to Z)\,\land\,(Z\to X)\,.$$
