Sets $A,B,C$ with $B\subseteq C$, prove that $(A-B)-C=A-C$ Ran across this and couldn't figure out how you would give a formal proof. It seems intuitive, in that $(A-B)-C$ is the elements in $A$ but not in $B$, and then also remove the elements from $(A-B)$ that are in $C$. But since $B$ is a subset of $C$, it almost feels like the subtraction of $B$ from $A$ is redundant, since they'd be removed when $C$ is subtracted anyway. Help would be appreciated, thanks!
 A: You are correct. Since $B \subseteq C$, when you remove $C$ from $A$, you are essentially removing $B$ as well. Here's the solution:
$\rightarrow$ Let $x \in (A-B)-C$.
So, $x \in (A-B)$ and $x \notin  C$.
So, $x \in A$ and $x \notin B$ and $x \notin C$.
So, $x \in A-C$. Thus, $(A-B)-C \subset A-C$.
$\leftarrow$ Now, let $x \in A-C$.
So, $x \in A$ and $x \notin C$.
Since $x \notin C$, this implies $x \notin B$. (Since $B \subseteq C$)
So, $x \in A$ and $x \notin B$ and $x \notin C$
So, $x \in A-B$ and $x \notin C$.
So, $x \in (A-B)-C$. Thus, $A-C \subset (A-B)-C$.
Therefore, $(A-B)-C = A-C$.
A: $$
(A-B)-C = (A\cap B^C)\cap C^C
= A\cap(B^C\cap C^C)
= A\cap(B\cup C)^C
= A\cap C^C
= A-C
$$
A: Let $x\in (A-B)-C$. Then $x \in (A-B)$ but $x\not\in C$. Thus $x\in A$ and $x\not\in B$. Since $x\in A$ and $x\not\in C$, $x\in A-C$. Thus $(A-B)-C\subseteq  A-C$.
Conversely, let $x\in A-C$. Then $x\in A$ and $x\not\in C$. By contrapositive of subset, $x\not\in B$. Thus $x\in (A-B)$. Also, since $x\not\in C$, $x\in (A-B)-C$. Thus $A-C\subseteq (A-B)-C$.
Putting both together gives the desired claim.
A: To provide an alternative view, here is a 'logical' proof: one where we expand the set theory definitions to the level of logic, and then use the laws of logic to simplify things.$
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}}
\newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} }
\newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
\newcommand{\ref}[1]{\text{(#1)}}
\newcommand{\then}{\Rightarrow}
\newcommand{\followsfrom}{\Leftarrow}
\newcommand{\true}{\text{true}}
\newcommand{\false}{\text{false}}
$
We are given that $\;B \subseteq C\;$, which means that for all $\;x\;$,
$$
\tag 0
x \in B \;\Rightarrow\; x \in C
$$
or alternatively
$$
\tag{0'}
x \in B \lor x \in C \;\equiv\; x \in C
$$

We now start at the most complex side, and calculate which elements $\;x\;$ it contains:
$$\calc
  x \in (A - B) - C
\op\equiv\hint{definition of $\;-\;$, twice}
  x \in A \land x \not\in B \land x \not\in C
\op\equiv\hint{DeMorgan -- to bring $\;B,C\;$ together, as in our assumption}
  x \in A \land \lnot (x \in B \lor x \in C)
\op\equiv\hint{using our assumption $\;B \subseteq C\;$, in the form $\ref{0'}$}
  x \in A \land \lnot (x \in C)
\op\equiv\hint{definition of $\;-\;$}
  x \in A - C
\endcalc$$
By set extensionality, this proves the original statement.
A: Assuming $x$ is an element in $A-C$, prove that it's an element in $(A-B)-C$
and vice versa.
