I am having trouble proving the theorem below:
$A \cup (B - C) = (A \cup B) - (C - A)$
Using the direct method I am to assume that the hypothesis is true. So that is where I start:
$\begin{array}{ccl} A \cup (B-C) & = & \{x \;|\; x\in A \vee x \in (B-C) \} \\ & = & \{x \;|\; x \in A \vee (x \in B \wedge x \not\in C)\} \\ & = & \{x \;|\; (x \in A \vee x \in B) \wedge (x \in A \vee x \not\in C)\} \end{array}$
But I don't know where to go from here. I have $(A \cup B)$ on the right side but how do I derive "$- (C - A)$"?