I want to prove the following statment: $$(A \cup C)\setminus B=(A\setminus B)\cup C \iff B\cap C= \varnothing$$ Do I need to prove each side? Or is one side enough? I mean, if I get from the left side to the right is it enough?
How do the following statements help me?
I would like to get some hint to prove this statement.
Thanks!
Hints:
Using $X \setminus Y = X \cap Y'$ and some simplifications you should be able to arrive at
$$(A \cup C) \cap B' \quad\supseteq\quad C$$
and then use $$X \cap Y = \varnothing \quad \iff \quad X \subseteq Y'$$ to show that $B \cap C = \varnothing$.
The other direction is quite strighforward, first expand everything and then use $$X \subseteq Y' \iff X \cap Y' = X.$$
I hope this helps ;-)
If $B\setminus C=\emptyset$, it is easy to check that $$ (A\cup C)\setminus B=(A\setminus B)\cup C. $$ Now for the other direction, one can use this fact that $$ A\subset B \text{ iff }A\cup B=B. $$ From $(A\cup C)\setminus B=(A\setminus B)\cup C$, one obtains $$ (A\setminus B)\cup (C\setminus B)=(A\setminus B)\cup [(C\setminus B)\cup(B\cap C)] $$ or $$ (A\setminus B)\cup (C\setminus B)=[(A\setminus B)\cup(C\setminus B)]\cup(B\cap C)]. $$ Using the fact, one can have $$ B\cap C\subset (A\setminus B)\cup (C\setminus B)=(A\cup C)\setminus B $$ from which one has $$ B\cap C=\emptyset. $$
If $X$ and $Y$ are sets, then $X\setminus Y=X\cap Y'$ by definition (where $Y'$ denotes the complement of $Y$ in some universal set). Also, if $X,Y,Z$ are sets, then $(X\cup Y)\cap Z=(X\cap Z)\cup (Y\cap Z)$ and $(X\cap Y)\cup Z=(X\cup Z)\cap (Y\cup Z)$(de Morgan's laws). If we use the first sentence of this answer, then we can compute:
(1) $(A\cup C)\setminus B=(A\cup C)\cap B'$, and
(2) $(A\setminus B)\cup C=(A\cap B')\cup C$.
Can you now use the second sentence of this answer (de Morgan's laws) to show that (1) and (2) are equal if and only if $B\cap C=\emptyset$? Hint: (1) and (2) can be rewritten, using de Morgan's laws, as:
(1') $(A\cup C)\setminus B=(A\cap B')\cup (C\cap B')$, and
(2') $(A\setminus B)\cup C=(A\cup C)\cap (B'\cup C)$.
Use the distributivity of $\cup$ and $\cap$ over each other (i.e, de Morgan's laws!) to rewrite (1') in a form similar to (2').
I hope this helps!
Let me try to explain how I'd think about this problem, since I'd work directly from definitions and thus use fewer formulas than other people seem to like. For the left-to-right implication, notice that $(A\setminus B)\cup C$ contains (by definition) all the elements of $C$. But it equals $(A\cup C)\setminus B$ and therefore contains none of the elements of $B$. So no elements of $C$ can be in $B$; that is, $B\cap C=\varnothing$.
For the right-to-left implication, look at the expressions $(A\cup C)\setminus B$ and $(A\setminus B)\cup C$ on the left side. Each involves starting with $A$, adding in the members of $C$ and removing the members of $B$; they differ only in the order in which "adding" and "removing" are done. So for points that are in only one of $B$ and $C$ or in neither of them, the two expressions work exactly the same way. A difference arises only for points in $B\cap C$ as these would be first removed and then added in $(A\setminus B)\cup C$, so they'd be present in the final result, but first added and then removed in $(A\cup C)\setminus B$, so they'd be absent from the final result. Conclusion: The two expressions on the left differ only in regard to elements of $B\cap C$. In particular, if $B\cap C=\varnothing$ then the two expressions agree and the equation on the left side therefore holds.
You can use an approach like in my answer to your other similar question. Let me know if you like a fuller answer, and I'll update.
A \setminus B. – dfeuer Jul 15 '13 at 15:22