Proving $(A \cup B) - (A \cup C) \subset A \cup (B - C)$

Question

I am trying to prove that $(A \cup B) - (A \cup C) \subset A \cup (B - C)$. Here is as far as I managed to get.

Let $x \in (A \cup B) - (A \cup C)$. Then: \begin{align*} x \in (A \cup B) - (A \cup C) & \implies x \in (A \cup B) \text{ and } x \not \in (A \cup C) \\ & \implies (x \in A \text{ or } x \in B) \text{ and } (x \not \in A \text{ and } x \not \in C) \\ & \implies ((x \in A \text{ or } x \in B) \text{ and } x \not \in A) \text{ and } x \not \in C \\ & \implies (x \in B \text{ and } x \not \in A) \text{ and } x \not \in C \\ & \implies x \in B - A \text{ and } x \not \in C \end{align*} In other words, I cannot managed to get there, and have gone down a wildly peculiar path. Any help to get me on the right track would be appreciated.

Answer 1

Once you conclude that

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

Consider two cases,

• if $x \in A$, then you are done.

• If $x \notin A$, then we must have $x \in B$ and $x \notin C$.

That if we have shown that $x \in A \cup (B-C).$

Answer 2

\begin{align}(A\cup B)\setminus(A\cup C)&=((A\cup B)\setminus A)\setminus C=(B\setminus A)\setminus C=\\&=(B\setminus C)\setminus A\subseteq B\setminus C\subseteq A\cup(B\setminus C)\end{align}

Answer 3

If $x \in A \implies x \in A\cup (B-C)$. If $x \notin A \implies x \in B$ and $ x \notin C \implies x \in (B-C) \implies x \in A\cup (B-C)$. You’re done.