A big problem is that I never even know where to start with proofs. Then I panic and get absolutely nowhere.
To reiterate: Prove that $(B – A) \cup (C – A) = (B \cup C) – A$ by showing that each side is a subset of the opposite side.
$$(B – A) \cup (C – A) = (B \cup C) – A$$
I thought to use the Distributive Laws, but I'm not sure if that would take me in the right direction.
I'm also supposed to prove it using membership tables, which I haven't even a basic understanding of. Blegh.
Edit: Here's a rough proof I attempted. How off the mark am I? And how can I rewrite it cleaner?
\ Let $x \in (B – A) \cup (C – A)$.
\ Then $x \in (B – A)$ or $x \in (C – A)$.
\ Assume that $x \in (B – A)$.
\ Thus $x \in B$ and $x \notin A$.
\ Therefore $x \in (B \cup C)$.
\ Because $x \notin A$, $x \in (B \cup C) – A$
\ $(B – A) \cup (C – A) \subseteq (B \cup C) – A. \quad \quad$ (Distributive Law)
\ If $x \in (B \cup C) – A$, then $x \notin A$, and $x \in B$ or $x \in C$.
\ Therefore $x \in (B – A)$ or $x \in (C-A)$
\ $x \in (B – A) \cup (C – A)$
\ $(B \cup C) – A \subseteq (B – A) \cup (C – A)$
Best Answer
Hint: pick an element from the left hand side, which has to be in either $B-A$ or $C-A$, and show it has to be in the right hand side. Then, pick an element from the right hand side and show it has to be in the left hand side.