How do I do this proof? $(A \cup B) - C = (A - C) \cup B \iff B \cap C = \emptyset$

Question

$(A \cup B) - C = (A - C) \cup B \iff B \cap C = \emptyset$

I know that for this problem, $C$ is removed from $A \cup B$

I also know that $(A - C) \cup B$ can be rewritten as $(A \cup B) - (C \cup B)$, but not sure what exactly that means. My assumption is that $A \cup B$ remains after removing $C \cup B$ leaving me with just $A \cup B$, but that doesn't seem to make sense if $C \cup B$ was removed. Am I just left with $A$ in this part?

I'm also a bit confused on how $B \cap C = \emptyset$

Would the proof then go as follows?

1) $(A \cup B) - C \subseteq (A - C) \cup B$

2) $(A - C) \cup B \subseteq (A \cup B) - C$

3) $B \cap C \subseteq \emptyset$

4) $\emptyset \subseteq B \cap C$

Would I then show that:

5) $(A \cup B) - C = (A - C) \cup B \Rightarrow B \cap C = \emptyset$

6) $B \cap C = \emptyset \Rightarrow (A \cup B) - C = (A - C) \cup B$

Am I on the right path? If so, how do I go about formulating each step? Seems like 6 different proofs in one is a bit much.

Answer 1

Edit:

Based on my answer in this post, I will give a new proof of this theorem. $$ (A\cup B)-C=(A-C)\cup B\iff B\cap C=\varnothing\tag1 $$ Suppose $f_L$ is the equation of the left side and $f_R$ is the equation of the right. Then by lemma $2$ and $3$ in my answer \begin{align} f_L\,&=\,((A\cup B-C)-(A-C)\cup B)\cup((A-C)\cup B-(A\cup B-C)) \\ \,&=\,((A-C)\cup (B-C)-(A-C)\cup B)\cup((A-C)\cup B-(A-C)\cup (B-C)) \\ \,&=\,\varnothing∪(B-(A-C)\cup (B-C))\tag2 \\ \,&=\,(B-(B-C))-(A-C)\tag3 \\ \,&=\,(B\cap C)\cap(A^c\cup C) \\ \,&=\,B\cap C\tag4 \\ \,&=\,f_R \end{align} Where

$(2):\,B-C\subset B\,$ and $\,A-C\subset (A-C)\cup (B-C)$

$(3):\,X-(Y\cup Z)=(X-Y)-Z$

$(4):\,(B\cap C)\subset(A^c\cup C)$ and by the law of absorption.

So by lemma $4$ in my answer, $(1)$ holds. We can see that this method proves the theorem completely in algebra and in one step.

Answer 2

The first implication: $$B \cap C = \emptyset \implies (A \cup B) - C = (A - C) \cup B $$ Is easy to prove, the problem is the other implication.

You can not prove that something belongs to empty. Therefore, the other demonstration is made by an indirect method.

The statement is equivalent to saying: $$B \cap C \not = \emptyset \implies (A \cup B) - C \not = (A - C) \cup B $$

Let $x\in(A \cup B)-C \implies x\in(A \cup B) \land x \not \in C \implies (x\in A \lor x\in B) \land x \not \in C$

$\implies (x\in A \land x\not \in C) \lor (x\in B \land x\not \in C) \implies (x\in(A-C))\lor (x\in (B-C)) $

$\implies x\in((A-C)\cup (B-C))$

But, $B-C \not = B$, because $B \cap C \not = \emptyset$, then $((A-C)\cup B) \not \subseteq ((A-C)\cup (B-C))$, by definition of equality of sets: $((A-C)\cup (B-C)) \not = ((A-C)\cup B).$ $\blacksquare$