How to Disprove $A-(B-C)=A-(B\cup C)$? For $A,B,C$ sets. I know I want to show that it is not the case that $A-(B-C)$ is not equal to $A-(B\cup C)$, I also know that the definition of set deference is x is an element in A but not in B if I have set A-B, but to disprove this is getting a bit difficult. 

You can also disprove the statement by using truth tables, $A$ having the  truth value $1$ (or TRUE) if $x \in A$ and $0$ (or FALSE) if not. The table for the union of sets $A$ and $B$, corresponds to the table of a logical OR (not exclusive), hence $A \vee B$ and set difference $A \setminus B$ corresponds to ($A$ AND (NOT 
$B$)), hence $A \wedge \neg B$. You can then disprove $A-(B-C)=A-(B\cup C)$ by converting your left hand side and right hand side into logical statements and showing that they have different truth tables. I hope this seems clear to you, if not, do not hesitate to ask questions!
