Is it sensible to look at intersection of events conditioned on complementary events? I am looking at a publication from a STEM-discipline (not mathematics), and I am little puzzled by an expression of the form
$$\{A \geq B \mid X = 1, Y = 1\} \cap \{A \leq B \mid X = 0, Y = 1\}$$
Here A, B, X, Y are discrete random variables with binary outcomes {0,1}. Therefore {X=0} and {X=1} are complementary events, so they cannot simultaneously occur. Can the above intersection be interpreted as anything other than the zero event?
 A: In standard probability theory, a probability may be conditional, but an event is not conditional; i.e., an expression like $P(A\mid B)$ does not refer to some object denoted $A\mid B\ $. Rather, $P(A\mid B)$ is parsed as "(the probability of $A$)(given $B$)", not as "(the probability of )($A$ given $B$)".  
On the other hand, in de Finetti's theory and similar non-standard theories, probability is defined in terms of such conditional events. In particular, any conditional event has one of three possible truth-values, where $A$ and $B$ are any ordinary unconditional events (that is, propositions):
$$A\mid B \quad\text{is}\quad\begin{cases}
\mathrm{True},  & \text{if }B\text{ is }\mathrm{True}\text{ and }A\text{ is }\mathrm{True} \\
\mathrm{False},  & \text{if }B\text{ is }\mathrm{True}\text{ and }A\text{ is }\mathrm{False} \\
\mathrm{Void},  & \text{if }B\text{ is }\mathrm{False}. 
\end{cases}$$ 
In these theories the conjunction of conditional events, say $\alpha$ and $\beta$, is defined as follows:
$$\alpha\,\And\,\beta\quad\text{is}\quad\begin{cases}
\mathrm{True},  & \text{if both $\alpha$ and $\beta$ are }\mathrm{True}\\
\mathrm{False},  & \text{if either $\alpha$ or $\beta$ is }\mathrm{False} \\
\mathrm{Void},  & \text{otherwise}. 
\end{cases}$$ 
Applied to your conditional event $\{A \geq B \mid X = 1, Y = 1\} \cap \{A \leq B \mid X = 0, Y = 1\}$ (assuming $\cap$ corresponds to $\And$), we find that it may be $\mathrm{False}$ (which I suppose corresponds to what you call the "zero event"), or it may be $\mathrm{Void}$.
NB: It can be shown (by just filling out the truth tables) that generally 
$$(P\mid Q)\,\And\,(R\mid S)\ \equiv\ PQRS\mid(\bar PQ\lor QS\lor S\bar R)
$$ 
where $\equiv$ indicates equality of truth-values.
