proof performed in inclusion of sets I want to prove the following identity:
$$(A\cup B)\cap (A\cup C)\subseteq A\cup (B\cap C)$$
From this equivalence I know that for a full test I should be able to prove that the left term is included into the right term and the other way around. However, I want to do only the first part of the proof I mean to  show that the left term is included into the right term.
So my proof is as follows:
Assume that we have $$x\in (A\cup B)\cap (A\cup C)$$
so we have two parts:
$$x\in (A\cup B)\simeq x \in A \vee x \in B \hspace{10 mm}(1)$$
and:
$$x\in (A\cup C) \simeq x \in A \vee x \in C \hspace{10 mm}(2)$$
From (1) I can assume that if we have an $ x\in B $ I could also say that there might exist a $x$ that also belongs to $C$; therefore:
$x\in B \wedge x\in C \hspace{10 mm}(3) $
From (3) and (1) I would have that:
$$ x\in A \vee (x \in B \wedge x \in C) \simeq  x \in A \cup(B\cap C)\hspace{10 mm}(4) $$
From (2) we have that $x\in A \vee x\in C$. If I consider that $x\in C$, I can also assume that there could exist a $x \in B \wedge x\in C \hspace{10 mm}$ (5).
Considering (2) and (5) I would end up having $x\in A \vee (x \in B \wedge x \in C)$, which is similar to: $x \in A \cup(B\cap C)$; finishing my proof.
The question that I have is if the proof that I have done is correct, or it is safe to convert the notation set to logical equivalences and then apply the distributive property. Any thoughts?
Thanks

Other responses have reacted to the OP's (i.e. original poster's) approach, and responded to his questions, concerning his approach.
For what it's worth, I would have approached the problem the same way that Karl suggested, in his comment.  Either $x \in A$ or $x \not\in A$.
Still another approach is to use a truth table to analogize a Venn Diagram.
\begin{array}{| r | r | r | r| r |}
\hline
  A & B & C & \text{LHS} ~= (A \cup B) \cap (A \cup C) & \text{RHS} ~= A \cup (B \cap C) \\
\hline
  T & T & T & T & T \\
  \hline
  T & T & F & T & T \\
  \hline
  T & F & T & T & T \\
  \hline
  T & F & F & T & T \\
  \hline
  F & T & T & T & T \\
  \hline
  F & T & F & F & F \\
  \hline
  F & F & T & F & F \\
  \hline
  F & F & F & F & F \\
  \hline
\end{array}
The assertion that LHS $~\subseteq~$ RHS is equivalent to the truth table assertion that for whatever rows where the $4$th column is true, that the $5$th column is also true.
For what it's worth, the above truth table actually demonstrates that LHS $~=~$ RHS, since you also have that for whatever rows the  $5$th column is true, that the $4$th column is also true.
