Set theory $A-(B-A) = A-B$ Determine which of he following statements are true for all sets $A,B,C,D$. If a double implications fails, determine whether one or the other statement of the possible implication holds, If an equality fails, determine whether the statement becomes true if the “equals” is related by one or the other of the inclusion symbols $\subseteq$  and left contains sign).
I was trying to solve this and stumbled upon a solution using the Universal set, but I don't know how to use the universal set. Is it possible to solve without it? This is what the first part of the solution looks like.
Q:  $A-(B-A) = A-B$
this statement is false, so I proceed to try the other statement: $A-(B-A) \subseteq A-B$:
$$A-(B-A) = A\cap(U - (B\cap(U - A)))$$
and he expanded this, but again I have no experience with the Universal set so can I show the statement is true without it?
Edit\attmept at $A-(B-A) \subseteq A-B$:
let $x\in A-(B-A)\implies x\in A$ or $x\notin B-A$  or $x\in A$ and ($x\in B\ and\ x\notin A)\implies$ but $x\in A$ in both case therefore $A-(B-A)\subseteq A-B$
 A: Instead of manipulating symbols, try drawing out a Venn diagram or working with explicit examples. This helps to build intuition with set theory (this can help to compensate for when our algebraic intuition with manipulating symbols fails us).
It turns out that the correct simplification is that $A - (B - A) = A$. In general, we can always think of $S - T$ as $S - (S \cap T)$, since $S - T$ is always a subset of $S$ (and so it's impossible to removing things from $S$ that don't even belong in $S$). Using this identity, we have:
$$
A - (B - A) = A - (A \cap (B - A)) = A - \varnothing = A
$$
A: It appears you are still struggling; general piece of advice: do not try to show a set equivalence (i.e., an identity for sets, as you have in your problem) by using an element-chasing proof unless 1) you are explicitly required to, or 2) a different approach is not very accessible.
Whenever you are confronted with an alleged identity and asked to declare whether or not it is true or false, in which case you are generally asked to supply a proof or give a counterexample, respectively, I would recommend first trying extreme examples (such as using empty sets and the like). If you cannot find a counterexample after a few tries, then that's fine--try to prove it. If you run into some major obstacles while trying to prove an identity for sets, then chances are your identity actually is not an identity (as in this case). Always try to use set algebra, as in the last part of Adriano's answer, before attempting an element-chasing proof. As you know, to prove an identity using an element-chasing proof, you must essentially give two proofs: 1) the first set is a subset of the second set, and 2) the second set is a subset of the first set (by mutual subset inclusion, then, the sets are equal). 
OK. That being said, I'll try to outline something below that should get you to the other side hopefully. First, note that the set difference denoted by $P-S$ is also often denoted by $P\setminus S$ or $P\cap S^C$, where $S^C$ denotes the complement of $S$ (and this may often be denoted by $\overline{S}$). The main point, in the context of what follows, is that $P-S$ and $P\cap S^C$ are equivalent by definition. See if you can follow this chain of equivalences:
\begin{align}
A-(B-A) &= A\cap(B-A)^C\tag{by definition}\\[0.5em]
&= A\cap(B\cap A^C)^C\tag{by definition}\\[0.5em]
&= A\cap(B^C\cup A)\tag{DeMorgan}\\[0.5em]
&= (A\cap B^C)\cup (A\cap A)\tag{distributivity}\\[0.5em]
&= (A-B)\cup A\tag{by definition}\\[0.5em]
&= A.
\end{align}
Does that make sense? Two things should now be clear:
First, by the argument above, we can see that $A-(B-A)=A$, hence $A-(B-A)\neq A-B$. 
Second, the above argument shows the reason behind Mehdi's comment, namely that $A-B\subseteq A-(B-A)$. How? Well, it is fairly easy to see that $A-B\subseteq A$, and this is why $(A-B)\cup A$ simplifies to $A$. However, we just established that $A-(B-A)=A$. Thus, if $A-B\subseteq A$, and $A = A-(B-A)$, then clearly $A-B\subseteq A-(B-A)$. That should probably be enough to answer your question (and none of it resorted to referring to the universal set).
A: Rather than just jumping in and trying to prove things by manipulating elements, or sets and unions and whatever else you can think of, try some examples!
The most "extreme" examples are usually easy to think of, and the most enlightening. Are either of those statements true if $A$ and $B$ are disjoint; $A \cap B = \varnothing$? How about if $A \subseteq B$? Or $B \subseteq A$? And rather than working generally, pick specific $A$ and $B$ like $A = \{1, 2\}, B = \{1, 2, 3, 4\}$. They don't need to be huge sets, or fancy, to give you a feeling about whether the statement is true or false.
Once you stumble on something that seems true, don't ignore the tried-and-true showing "both sets contain each other" to show set equality. That is, to show that $X = Y$ given two sets $X, Y$, it's often straightforward to show that $X \subseteq Y$ and $Y \subseteq X$; so if you pick $x \in X$, show $x \in Y$ to show that $X \subseteq Y$, for example.
A: Hint: Use $A-B = A \cap B'$ and De Morgan's law. Use Venn diagram to visualize what is happening.
A: $A-(B-A)=A\cap(\overline{B\cap \overline{A}})\stackrel{\text{DM}}=A\cap (\overline{B}\cup A)=\text{LHS}$.   
$x\in A,\ x\in B\,\Rightarrow\, x\in\text{LHS}$, yet $x\not\in \text{RHS}=A-B=A\cap\overline{B}$.   
DM-DeMorgan.
