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If $A$, $B$, and $C$ are sets and $B \subseteq C$, how would I be able to show that $(A - B) - C = A - C$? I understand how this would be trivial if the set $B$ has no elements (or in other words its basically empty), but I'm confused as to how would you go about proving this otherwise?

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  1. Suppose $x\in(A-B)-C$. Then we know $x\in A-B$ and $x\notin C$. From $x\in A-B$ we know $x\in A$ and $x\notin B$. Since we know $x\in A$ and $x\notin C$, we know $x\in A-C$.

  2. Suppose $x\in A-C$. Then we know $x\in A$ and $x\notin C$. Since $x\notin C$ and $B\subseteq C$, we know $x\notin B$. Then since $x\in A$ and $x\notin B$, we know $x\in A-B$. Since $x\in A-B$ and $x \notin C$, we know $x\in (A-B)-C$.

From (1), we know $(A-B)-C \subseteq A-C$.

From (2), we know $A-C \subseteq (A-B)-C$.

Therefore, $(A-B)-C = A-C$.

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$$\begin{align}&x\in(A\setminus B)\setminus C\\ \iff &x\in A\setminus B\land x\notin C\\ \iff &x\in A\land x\notin B\land x\notin C\end{align}$$ From $B\subseteq C$ we have $x\in B\rightarrow x\in C$ or equivalently via propositional logic $x\notin C\rightarrow x\notin B$ and $x\notin C\leftrightarrow (x\notin B\land x \notin C)$. Hence ultimately $$\begin{align}&x\in(A\setminus B)\setminus C\\ \iff &x\in A\land x\notin C\\ \iff &x\in A\setminus C \end{align}$$

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In elementary problems of this type you should think first of proving the equality of the two sets by showing that each is a subset of the other: $(A\setminus B)\setminus C\subseteq A\setminus C$ and $A\setminus C\subseteq(A\setminus B)\setminus C$. Very often the easiest way to do this is by ‘element-chasing’: to show that $X\subseteq Y$, let $x$ be an arbitrary element of $X$ and show somehow that this forces $x$ to belong to $Y$.

In your problem, to show that $(A\setminus B)\setminus C\subseteq A\setminus C$ you might start like this:

Let $x\in(A\setminus B)\setminus C$ be arbitrary; we wish to show that $x\in A\setminus C$, i.e., that $x\in A$ and $x\notin C$. By hypothesis $x\in A\setminus B$ and $x\notin C$, so all that remains is to show that $x\in A$. But that’s clear: $x\in A\setminus B$, so $x\in A$ and $x\notin B$. We don’t care that $x\notin B$: we now know that $x\in A$ and $x\notin C$, and hence that $x\in A\setminus C$, as desired.

I’ll leave the other inclusion to you; try to follow the same general scheme, and don’t be afraid to use words as well as symbols, just as I did above. The idea is to produce an argument that is both convincing and clear to the reader.

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