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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.

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    I also know I must show this element wise for sure, just stuck on how to show that A-(B-C) is not a subset of A-(BUC) – 2017-02-06
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    Try a few examples. That's how to disprove something like this. Hint: What if $B=C$? – 2017-02-06
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    Draw a few Venn diagrams with various relationships between B and C, and see what works and what fails. – 2017-02-06
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    Sweet I see it with counter example, so basically from what I understand if A={1,2,3} and B=C={1,2} then I will show that {1}=A-(B-C) and that {1,3}= A-(BUC) therefore {1} is an element of A-{B-C} and {1,3} is an element in A-(BUC) thus A-(B-C) is not a subset of A-(BUC)? – 2017-02-06
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    That should work, but note you could also have situations where the "not a subset" goes in the other direction. – 2017-02-06
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    but if I have to disprove a bi-conditional shouldn't it be sufficient to just show one side is not equal to the other? I believe this proof reads like an iff proof so I thought just proving one side should be okay? – 2017-02-06

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Basic idea of the proof goes as follows: Let $x \in A \cap C$ and $x \notin B$, then $x \in A-(B-C)$ but $x \notin A-(B \cup C)$. Hence, the two sets are not equal.

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    Thank you so much these are very helpful. I appreciate everyone for your help. The diagramming made me a bit confused for like a few minutes but now it makes sense to after your explanation. – 2017-02-06
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We can disprove this by simply providing a counterexample. Let $A=(0,1)$, let $B=(0,1/2)$ and let $C=(0,1/2)$.

Just in case you don't know, for $a

Anyways, $A-(B-C)=(0,1)$ while $A-(B\cup C)=[1/2,1)^\dagger$ and clearly $(0,1)\ne [1/2,1)$. And so, this is the requisite counterexample to the statement.

$\dagger: [1/2,1):=\{x\in \mathbb{R}: 1/2\le x<1\}$.

NB: This is just a single example. This would work for $A=\{1,2,3\}$ with $B=C=\{1,2\}$.

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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!

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Suppose $B$ is a non-empty subset of $A$ and $B=C.$ Then the LHS is $A$ and the RHS is $A$ \ $B$ which is not equal to $A.$ So the equation cannot be true for all $A,B,C.$

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While the other answers already gave counterexamples, and while truth tables will always point you to sufficient counterexamples, the following is more about how to "look for" easy counterexamples.

I want to show that it is not the case that $A-(B-C)$ is not equal to $A-(B\cup C)$

The difference between the two expressions is $\;B \setminus C\;$ vs. $\;B \cup C\;$. The former is a (smaller) subset of $B$ since $\;B \setminus C \subseteq C\;$, while the latter is a (larger) superset of $B$ since $B \cup C \supseteq B\,$. When does this difference get exacerbated the most? When $B$ is the "smallest" possible, so consider $B = \emptyset\,$.

For $B=\emptyset$ obviously $\;B \setminus C = \emptyset\;$ and $B \cup C = C\,$, so $A \setminus (B \setminus C) = A$ and $A \setminus (B \cup C)=A \setminus C$. Again, look at the difference between the two sides, it's the $\setminus C$ on the RHS. When does that difference get exacerbated the most? When $C$ is the "largest" possible, and since we are only interested in the relation to $A$, that means $C=A$.

So in the end, one natural counterexample is $B=\emptyset, C=A$ for which:

$$A \setminus (B \setminus C) = A \setminus (\emptyset \setminus A) = A \setminus \emptyset = \;A \quad\ne\quad \emptyset\; = A \setminus A = A \setminus(\emptyset \cup A) = A \setminus (B \cup C) $$

(Note: the latter holds true for $A = \emptyset$ of course, but the question was about arbitrary sets, so this still is a sufficient counterexample.)