# Using distributive laws to manipulate a union and intersection of sets

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.

-
Can you see why the two stated inequalities are intuitively obvious, or at least seem like they ought to be true? – Ben Millwood Sep 17 '12 at 1:14
Not entirely. The first equation only makes sense if $B-A = B$ which is only true if B and A are disjointed (I think). The second equation only makes sense if $A \cap B = A$ which is only true if A is a subset of B. – rioneye Sep 17 '12 at 3:22
the best way to think of set equalities is "$x$ is in the LHS if and only if it's in the RHS". In the first one, we have "$x$ is in $A$ or $B$" versus "$x$ is in $A$, or it is in $B$ and not $A$". Using the "inclusive or" that allows for both things to be true, can you see why those two statements are equivalent? – Ben Millwood Sep 17 '12 at 14:07

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 $B-A=(B \cap A^c)=B \cap (A^c\cup B^c)=B\cap (A \cap B)^c=B-(A\cap B)$.

-
What theorem or definition says that $B-A=(B\cap CA)$? – rioneye Sep 16 '12 at 23:26
That $B-A=B \cap A^c$ is a general fact, you can convince yourself it is true by drawing a Venn diagram. – Tarnation Sep 16 '12 at 23:28
What about this step: $B \cap CA = B \cap (CA \cup CB)$? That doesn't seem very intuitive. – rioneye Sep 16 '12 at 23:43
Notice that $B \cap B^c= \emptyset$, so $B \cap A^c= B \cap (A^c \cup B^c)$. – Tarnation Sep 16 '12 at 23:46

$A, B \subset X$, 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

$A \cup (B - A) = A \cup (B \cap A^c)$

by distributivity

$= (A \cup B) \cap (A \cup A^c) = (A \cup B) \cap X = A \cup B$

-

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, $(A\cap B)^c=A^c\cup B^c$.)

-