= Solution
Using <set difference> as intersection with a <complement of a set>, then <De Morgan's laws> and distribution, gives
$$
\begin{aligned}
(A\cup B)\setminus(C\setminus A)
&=(A\cup B)\cap(C\cap A^c)^c\\
&=(A\cup B)\cap(C^c\cup A)\\
&=A\cup(B\cap C^c)\\
&=\boxed{A\cup(B\setminus C)}.
\end{aligned}
$$
For the third equality, expand the two <set unions>: the terms $A\cap C^c$ and $A\cap B$ are absorbed by $A$, and $A\cap A=A$. This proves the required identity entirely by <set> algebra.
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