Show that for any three sets A, B, C we have that
$(A ∪ B) − C = (A − C) ∪ (B − C)$
I proved like the following:
Let $x ∈ (A ∪B )-C$
$x ∈ A $ and $x ∈ B$ and $x ∉ C $
Let $y ∈ (A − C) ∪ (B − C)$
$y ∈ A $ and $y ∈ B$ and $y ∉ C $
Since $x = y$
$(A ∪ B) − C = (A − C) ∪ (B − C)$
Is this a valid proof?
Here is the forward direction. I'll leave the reverse direction for you to do.
Let $x\in (A\cup B)\setminus C$.
This means by definition that $x\in (A\cup B)$ and $x\notin C$.
By rewriting the first condition further, this means that [$x\in A$ or $x\in B$] while at the same time $x\notin C$.
From here, suppose without loss of generality that $x\in A$. (It could have instead been the case that $x\in B$ was the true statement and not $x\in A$, but the arguments that follow would be effectively the same, just replacing the letters $A$ for $B$ where appropriate). We have then $x\in A$ and $x\notin C$.
Then by the definition of set difference, this means that $x\in (A\setminus C)$.
It follows then that $x\in (A\setminus C) \cup (B\setminus C)$.
Since we have shown that $x\in (A\cup B)\setminus C$ implies that $x\in (A\setminus C)\cup (B\setminus C)$ this effectively shows that $(A\cup B)\setminus C \subseteq (A\setminus C)\cup (B\setminus C)$
The reverse direction works similarly. You may want to break into three cases however rather than just two: 1) $x$ is in $A$ but not $B$, 2) $x$ is in $B$ but not $A$, 3) $x$ is in both $A$ and $B$. Finally, remember that $\cup$ is similar to the logical or while $\cap$ is similar to the logical and. Don't mix them up.