I have the following distributive laws for some X,Y,Z sets: $X \cup (Y\cap Z) = (X \cup Y)\cap(X \cup Z)$ and $X \cap (Y\cup Z) = (X \cap Y)\cup(X \cap Z)$, with that I need to show that $A \cup B = A \cup(B-A)$ and $B-A = B - (A \cap B)$ for two sets $A,B \epsilon R$. I've been looking at this for about an hour but I'm not a hundred percent sure how to get from point A to B. The closest I've come to solving this is if I somehow argue that $A \cap B = A - B$ and $A \cup B = A + B$. I think I can prove the latter by saying that if A and B are disjoint then the equality holds, but I don't know how I could prove the former. Any insight would be appreciated.
For the first part, note that $A \cup (B-A)=A\cup(B\cap A^c)$, which by your distributive law equals $(A \cup B)\cap(A\cup A^c)=A\cup B$.
For the second part, we have $B-A=(B \cap A^c)=B \cap (A^c\cup B^c)=B\cap (A \cap B)^c=B-(A\cap B)$.
$A, B \subset X$, where $X$ is some set containing both $A$ and $B$ (for instance $X = A \cup B$). Note that $B - A = B \cap A^c$. Then
$A \cup (B - A) = A \cup (B \cap A^c)$
$= (A \cup B) \cap (A \cup A^c) = (A \cup B) \cap X = A \cup B$
Use the fact that $B\setminus A=B\cap(\Bbb R\setminus A)$: you appear to be working in $\Bbb R$, so $$A\cup(B\setminus A)=A\cup\Big(B\cap(\Bbb R\setminus A)\Big)=(A\cup B)\cap\Big(A\cup(\Bbb R\setminus A)\Big)=\dots\;?$$
It’s not immediately clear how you’re to use the distributive laws to show that $B\setminus A=B\setminus(A\cap B)$. This is equivalent to $B\cap(\Bbb R\setminus A)=B\cap\big(\Bbb R\setminus(A\cap B)\big)$, but that’s still not something to which you can apply the distributive laws. What you need is the distributive law $$X\setminus(Y\cap Z)=(X\setminus Y)\cup(X\setminus Z)\;,$$ in particular the case
$$\Bbb R\setminus(A\cap B)=(\Bbb R\setminus A)\cup(\Bbb R\setminus B)\;;$$
have you by any chance proved that yet? (You might have expressed it in terms of complements, $(A\cap B)^c=A^c\cup B^c$.)