Prove $B ⊆ (C ∪ A) ⇔ (B \setminus A) ⊆ C$ Prove the following:
\begin{align}
B ⊆ (C ∪ A) &⇒ (B\setminus A) ⊆ C \\
(B\setminus A) ⊆ C &⇒ B ⊆ (C ∪ A)
\end{align}
Using Eulerian circles I only understood that statements are true. Still have no idea how to prove. 
Any hints guys? (not asking for complete solution, need just an idea to start with). Would appreciate any help. :-)
 A: Hint: $"\implies"$ Assume that you know that  $B \subseteq (C \cup A)$. Take $x \in B \backslash A$. Then by definition $x \in B$ and $x \notin A$. But since $B \subseteq (C \cup A)$ then it must be that $x \in C$ (since $x \notin A$). This implies that for any $x \in B\backslash A$ it must be true that $x \in C$ which proves that $$B\backslash A \subseteq C$$ 
For the other direction assume that the RHS holds true and work similarly, i.e. take an element of $B$ and try to show that it must necessary be in $C \cup A$.   
A: Here is an alternative approach, which goes back to the definitions and then uses the laws of logic.
$
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\newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
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$For the first expression $\;B \;\subseteq\; C \cup A\;$, we get
$$\calc
B \;\subseteq\; C \cup A
\op\equiv\hint{definition of $\;\subseteq\;$; definition of $\;\cup\;$}
\langle \forall x :: x \in B \;\then\; x \in C \lor x \in A \rangle
\op\equiv
  \hints{logic: write $\;P \then Q\;$ as $\;\lnot P \lor Q\;$}
  \hint{-- usually it is a bit more difficult to manipulate $\;\then\;$}
\langle \forall x :: x \not\in B \lor x \in C \lor x \in A \rangle
\endcalc$$
Now do the same thing for the other expression $\;B \setminus A \;\subseteq\; C\;$, apply DeMorgan, and compare the results.
