Suppose $A$, $B$, and $C$ are sets, and $A - B \subseteq C$. Then $A - C \subseteq B$. I know how to prove it by contradiction, but I am wondering if it's possible to prove it directly. I tried doing that, but so far no results. Is it not possible to prove it directly?
Thanks.
 A: We have
$$A-B=A\cap B^c\subset C\Rightarrow C^c\subset A^c\cup B$$
hence
$$A-C=A\cap C^c\subset A\cap  (A^c\cup B)=A\cap B\subset B$$
A: Let $a\in A-C$, then $a\in A$ and $a\notin C$. Therefore $a\notin A-B$. So either $a\notin A$ or $a\in B$. Since $a$ Was taken from $A$, we have that $a\in C$ as wanted.
A: $
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\calcop}[2]{\notag \\ #1 \quad & \quad \text{"#2"} \notag \\ \quad & }
\newcommand{\endcalc}{\notag \end{align}}
$I would use the definitions to calculate what it means that $\;A - B \subseteq C\;$, as follows:
$$\calc
A - B \subseteq C
\calcop{\equiv}{definition of $\;\subseteq\;$; definition of $\;-\;$}
\langle \forall x :: x \in A \land x \not\in B \;\Rightarrow\; x \in C \rangle
\calcop{\equiv}{logic: write $\;P \Rightarrow Q\;$ as $\;\lnot P \lor Q\;$; DeMorgan}
\langle \forall x :: x \not\in A \lor x \in B \;\lor\; x \in C \rangle
\endcalc$$
Since the last line is symmetric in $\;B\;$ and $\;C\;$, the first line is as well, which proves
$$
A - B \subseteq C \;\equiv\; A - C \subseteq B
$$
as requested.
A: There are many forms to do directly: Since $A-B\subset C$ then $(A\cap \overline{B})-C=\emptyset$ wich is the same as $A\cap \overline{B}\cap \overline{C}=\emptyset$. Now what you want to prove is precisely this. Convince your self.
A: We know that $A ⊆ B ∪ C ↔ A - C ⊆ B$ and $A ∪ B = B ∪ A$ so
$$A − B ⊆ C ↔ A ⊆ C ∪ B ↔ A ⊆ B ∪ C ↔ A − C ⊆ B.$$
