Simple Logic Question I've very little understanding in logic, how can I simply show that this is true:
$$((X \wedge \neg Y)\Rightarrow \neg Z) \Leftrightarrow ((X\wedge Z)\Rightarrow Y)$$
Thanks a lot.

You want to show that
$$((X \wedge \neg Y)\Rightarrow \neg Z) \Leftrightarrow ((X\wedge Z)\Rightarrow Y).$$
It is hard to know without context what "show" might mean.  For example, we could be working with a specific set of axioms. Since an axiom system was not specified, I will assume we are looking for a precise but not axiom-based  argument.
Truth tables are nicely mechanical, so they are a very good way to verify the assertion. Below we give a "rhetorical" version of the truth table argument. It probably shows that truth tables would have been a better choice! However, it is important to be able to scan a sentence and understand under what conditions that sentence is true.
We want to show that (a) if $(X \wedge \neg Y)\Rightarrow \neg Z$ is true then   $(X\wedge Z)\Rightarrow Y$ is true and (b) if $(X\wedge Z)\Rightarrow Y$ is true then $(X \wedge \neg Y)\Rightarrow \neg Z$ is true.
We deal with (a). There are two ways for $(X \wedge \neg Y)\Rightarrow \neg Z$ to be true: (i) if $\neg Z$ is true or (ii) if  $X \wedge \neg Y$ is false. In case (i), $Z$ is false, which implies that $X\wedge Z$ is false, which implies that
$(X\wedge Z)\Rightarrow Y$ is true.  In case (ii), $X$ is false or $Y$ is true. If $X$ is false, then $X\land Z$ is false, and as in case (i), $(X\wedge Z)\Rightarrow Y$ is true. If $Y$ is true, then automatically $(X\wedge Z)\Rightarrow Y$ is true. We now have completed proving (a).
The proof for the direction (b) is very similar.
Another way: We can also use (Boolean) algebraic manipulation to show that each side is logically equivalent to $(\neg X \lor \neg Z)\lor Y$.
Note that
$(X \wedge \neg Y)\Rightarrow \neg Z$ is equivalent to $\neg(X\wedge \neg Y)\lor \neg Z$, which is equivalent to $(\neg X \lor Y)\lor \neg Z$, which is equivalent to $(\neg X \lor \neg Z)\lor Y$.
Note also that $(X\wedge Z)\Rightarrow Y$ is equivalent to $\neg(X\wedge Z)\lor Y$, which is equivalent to $(\neg X\lor \neg Z)\lor Y$.  This completes the argument.
