Prove that $A - (B \cap C) = (A - B) \cup (A - C)$

Question

Prove that $A - (B \cap C) = (A - B) \cup (A - C)$

How do I go about solving this proof? I've tried a few things but I'm stuck. I don't need the exact proof step by step, just a few hints on how to solve it would really help me.

edit: I tried to solve considering $3$ cases.

If $x$ is an element of $A−(B\cap C)$, then $x$ belongs to $A$ and $x$ doesn't belong to $B\cap C$.

Then, there I considered $3$ possibilites:

• $x\in A$ and ($x\in B$ or $x\in C$)

• $x\in A$ and ($x\notin B$ or $x\in C$)

• $x\in A$ and ($x\in B$ or $x\notin C$)

After that I found that all of these cases satisfied the definition of set difference which states that $A-B$ is equal to the elements of $A$ that are not in $B$. In that case my $B$ is "($B\cap C$)". After that I tried to prove the other side "$(A−B)\cup (A−C)$" but I got lost and didn't know what to do.

Answer 1

Let $A, B, C$ be sets and $x \in A \setminus (B \cap C)$.

$x \in A \setminus (B \cap C)$

$\Leftrightarrow x \in A \wedge x \notin (B \cap C)$

$\Leftrightarrow x \in A \wedge (x \notin B \vee x \notin C)$

$\Leftrightarrow (x \in A \wedge x \notin B) \vee (x \in A \wedge x \notin C)$

$\Leftrightarrow x \in (A \setminus B) \cup (A \setminus C)$.

The equivalencies give us that the LHS and the RHS are subsets of each other, so they are equal.

Answer 2

Hint To prove an equality of sets $X =Y$ you need to show that $X \subseteq Y$ and $Y \subseteq X$, where $X,Y$ are the expressions in that particular problem.

Hint 2 To show that $X \subseteq Y$ you start with "Let $x \in X$ be arbitrary." You explain what this means and try to get that $x \in Y$.