How to prove $(A \cup (B - C)) \cap B^c \subseteq A - B$

Question

My try:

Suppose $(A \cup (B - C)) \cap B ^c$. We know $x \in A \cup (B - C)$ and $x \in B^c$. So, $x \in A$ or $x \in B - C$ and $x \in B^c$.

From there, I see two possible cases: 1 - $x \in A$ and $x \in B^c$. This leads to $x \in A - B$ 2 - $x \in B - C$ and $x \in B^c$. Don't see how it's even possible to get into $x \in A - B$.

Maybe I don't need to divide into two steps.

Answer 1

From there, I see two possible cases:

1. $x \in A$ and $x \in B^c$. This leads to $x \in A - B$
2. $x \in B - C$ and $x \in B^c$. Don't see how it's even possible to get into $x \in A - B$.

Maybe I don't need to divide into two steps.

In the second case: ... $(B\smallsetminus C)\cap B^\complement$ is the empty set, which is a subset of any set, including $(A\smallsetminus B)$.

Answer 2

Consider left-hand side of the formula.

\begin{align} & (A\cup(B-C))\cap B^c &\\ & (A\cap B^c)\cup((B-C)\cap B^c) && \text{De Morgan's law}\\ & (A-B)\cup((B-C)-B) && \text{show that the second arg. to union is the empty set}\\ & (A-B)\cup\emptyset \\ & (A-B) \end{align} In fact you can put an equality relation sign between them.