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
 A: 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.
A: Some preparation helps in a purely symbolic proof so as to not get lost in details. For assertions $a,b$ let us write $a\cdot b$ for $a\land b,$ and $a+b$ for $a\lor b.$ Then "$\cdot$" and "$+$" are associative and commutative, and each is distributive over the other. Also $a\iff a\cdot a \iff a+(a\cdot c) \iff a+(b\cdot a).$
Let $a\iff x\in A$ and $b\iff x\in B$ and $c\iff x\in C.$ Then for any $x$ we have $$x\in (A\cup B)\cap (A\cup C)\iff (a+b)\cdot (a+c)\iff$$ $$  ((a\cdot a)+(a\cdot c)+(b\cdot a)+(b\cdot c))\iff$$ $$((a+a\cdot c)+(b\cdot a)+(b\cdot c))\iff $$ $$((a+b\cdot a)+(b\cdot c))\iff $$ $$(a+(b\cdot c))\iff$$ $$[x\in A\lor (x\in B\land x\in C)]\iff$$ $$ 
[x\in A\lor (x\in B\cap C)]\iff x\in A\cup (B\cap C).$$
