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So to find the simplest form I made an arbitrary sets A,B,C and then put them into the expression.

In the end I believe this simplifies to just C and i'm fairly confident this is the correct answer.

I was hoping you could offer a more correct or common way of simplifying this.

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3 Answers 3

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It should be just $C$. To see this, notice that $(A\cup B) \cap C$ means the element in C and also in A or B, and $ (C-A)-B $ means the element in C but not in A or in B. Thus the union of the previous statement means every element in C (in A or B, or not in A or B).

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You can write $$((C-A)-B) = C - (A \cup B)$$ and hence obtain that your original expression is equivalent to $$ (C - (A \cup B)) \cup (C \cap (A \cup B).$$ If you think about it for a moment, you'll realize that this is of the form $$ (C-D) \cup (C \cap D),$$ where $D = A \cup B,$ and this is just $C$ as you predicted.

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Let $D=((A \cup B) \cap C) \cup ((C \setminus A) \setminus B)$. We want to prove $x \in C \iff x \in D.$

Assume $x \in C$. If $x \in A \cup B$, then $x \in (A \cup B) \cap C$ so $x \in D$.

If $x \in A \cup B$, then $x \notin A$ and $x \notin B$, so $x \in C \setminus A$ and also $x \in (C \setminus A) \setminus B$, so again $x \in D$.

Conversely, if $x \notin C$, then $x$ is not in either component of the union that comprises $D$ so $x \notin D$ and we are done.

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