# Careful proof of $A \subseteq (B \cup C) \iff A \setminus C \subseteq B$

Carefully prove that:

If $A,B,C$ are non-empty sets then $A \subseteq (B \cup C) \iff A \setminus C \subseteq B$.

So i need to prove

$A\subseteq(B\cup C) \implies (A\setminus C)\subseteq B$

and then

$(A\setminus C)\subseteq B \implies A\subseteq(B\cup C)$

So $x\in A$..

Can you provide the step by step proof please. Thank you in advance

• There are usually no "step-by-step" answers in this site. Askers are expected to do some self effort on their own and complete to a whole answer. – DonAntonio Aug 30 '13 at 7:56
• The statement is true even if one or more of the sets, $A,\,B$ and $C$ are empty. – Alraxite Aug 31 '13 at 0:50

Suppose $A\subseteq B\cup C$. Let $x\in A\setminus C$. Then $x\in A$ and from $A\subseteq B\cup C$, we conclude $x\in B\cup C$. Now use the fact that $x\not\in C$.
To prove the reverse implication, suppose $A\setminus C\subseteq B$ and let $x\in A$. Now consider two cases: $x\in C$ and $x\not\in C$.
For the left hand side, \begin{align} & A \subseteq B \cup C \\ \equiv & \;\;\;\;\;\text{"definition of $\;\subseteq\;$"} \\ & \langle \forall x :: x \in A \;\Rightarrow\; x \in B \cup C \rangle \\ \equiv & \;\;\;\;\;\text{"definition of $\;\cup\;$"} \\ & \langle \forall x :: x \in A \;\Rightarrow\; x \in B \lor x \in C \rangle \\ \equiv & \;\;\;\;\;\text{"rewrite $\;\Rightarrow\;$"} \\ & \langle \forall x :: x \not\in A \lor x \in B \lor x \in C \rangle \\ \end{align}