How to prove that $A - (A \cap B) = A - B$ I've found a very similar question, but couldn't follow the only answer. I don't know how to get this: $A\cap(A^c\cup B)=(A\cap A^c)\cup (A\cap B)$, even if I see that they're equivalent.
I want to know how to prove this using set properties as $A \cap A' = 0$, $(A \cap B)\cap C = A \cap    (B \cap C)$, $A \cap \varnothing = \varnothing$, $A \cup \varnothing = A$,  if that's possible.
I can usually think with Venn diagrams and get the answer, especially because of the visual intuition, but they can't be used to prove anything and they won't get me far anyway. I understand that $A - (A \cap B) = A - B$, my problem is in proving it.
 A: It can be done with the algebraic properties of set operations, but I think the simplest, most self-evident proof is to just take an unknown element of each side, and show that it's an element of the other side.

Here's a proof based on that approach . . .

First we show the inclusion $A - (A \cap B) \subseteq A - B$.


*

*Let $x\in A - (A \cap B)$.$\\[4pt]$

*Then $x\in A$, but $x\notin A\cap B$.$\\[4pt]$

*Since $x\notin A\cap B$, $x$ is not in both $A$ and $B$.$\\[4pt]$

*Hence, since $x\in A$, it follows that $x\notin B$.$\\[4pt]$

*Thus, $x\in A$, but $x\notin B$, so $x\in  A - B$.


It follows $A - (A \cap B) \subseteq A - B$.

Next we show the inclusion $A - B\subseteq A - (A \cap B)$.


*

*Let $x\in A - B$.$\\[4pt]$

*Then $x\in A$, but $x\notin B$.$\\[4pt]$

*Since $x\notin B$, $x\notin A\cap B$.$\\[4pt]$

*Thus, $x\in A$, but $x\notin A\cap B$, so $x\in  A - (A\cap B)$.$\\[4pt]$


It follows $A - B\subseteq A - (A \cap B)$. 

Since we have both inclusions, it follows that $A - (A \cap B)=A - B$, as was to be shown.
A: Here is an algebraic like proof you seemed to want
instead of a point by point proof.  
A - (A $\cap$ B) = A $\cap (A \cap B)^c$ =
$A \cap (A^c \cup B^c)$ =
$(A \cap A^c) \cup (A \cap B^c) = A \cap B^c = A - B.$  
Notice the use of DeMorgan's rules for sets.
