Why does A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)? If I have the elements from sets A and B, and I want to find the set A ∪ (B ∩ C), I end up with just the elements of A. On the other hand, if I have the elements from A and B and want to find (A ∪ B) ∩ (A ∪ C), it seems like I end up with (A ∩ B) ∩ A, which is just (A ∩ B). This set was not a possible outcome with A ∪ (B ∩ C). 
Where has my reasoning going wrong? 
 A: Since $B \cap C \subseteq B$ and $B \cap C \subseteq C$, we have
$$A \cup (B \cap C) \subseteq A \cup B$$
and
$$A \cup (B \cap C) \subseteq A \cup C$$
This shows that $A \cup (B \cap C)$ is contained in both $A \cup B$ and $A \cup C$, so it is contained in their intersection:
$$A \cup (B \cap C)\subseteq (A \cup B) \cap (A \cup C)$$
This proves containment in one direction.
For the opposite direction, suppose that $x \in (A \cup B) \cap (A \cup C)$. There are two possibilities: either $x \in A$ or $x \not\in A$.
If $x \in A$ then certainly $x \in A \cup (B \cap C)$.
On the other hand, if $x \not\in A$, then $x$ must be in both $B$ and $C$, since $x \in (A\cup B) \cap (A\cup C)$. Consequently, $x \in B \cap C$, and therefore $x \in A \cup (B \cap C)$.
In both cases we have $x \in A \cup (B \cap C)$. This proves the containment
$$(A \cup B) \cap (A \cup C) \subseteq A \cup (B \cap C)$$
so we're done.
A: A ∪ (B ∩ C) means $ x : x \in A  $  or  $ x \in B,C $ or $ x \in A,B,C $
(A ∪ B) ∩ (A ∪ C) means $ x: x \in A $ or $ x \in B $ or $ x \in A,B $ and $ x \in A $ or $ x \in C $ or $ x \in A,C $ => look at this statement and see the possible combinations => $ x : x \in A  $ or $  x \in B,C $ or $ x \in A,B,C $. 
A: 
If I have the elements from sets A and B, and I want to find the set A ∪ (B ∩ C), I end up with just the elements of A.

Not generally, and more importantly: not relevant.
$\cup$ means union: $A\cup B$ is set of elements in either set A or set B.
$\cap$ means intersection: $B\cap C$ is set of elements in both set B and set C.

$A\cup(B\cap C) \subseteq (A\cap B)\cup(A\cap C)$   If you have an element either from set A or from both sets B and C,  then you have elements which are from both either sets A or B and from either sets A or C.

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*Can you show that this is so?   Hint: Use case work: (If an element is from set A then..., and if an element is from both B and C, then ...., therefore ....)

$A\cup(B\cap C) \supseteq (A\cap B)\cup(A\cap C)$   If you have elements which are from both either sets A or B and from either sets A or C, then you have elements which are either from set A or from sets B and C.

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*Can you show this too?

A: Let's think it out slowly.
Let $x \in A \cup B\cap C$.  Then either Case 1: $x \in A$ or Case 2: $x \in B \cap C$.  (or case 3: both).
Case 1: $x \in A$.  
So $x \in A \cup B$.  And $x \in A \cup C$.  So $x \in (A\cup B) \cap (A \cup B)$.
Case 2: $x \in B \cap C$.  
So $x \in B$.  So $x \in A \cup B$.
But $x \in C$ also.  So $x \in A \cup C$.
So $x \in (A\cup B)\cap (A \cup C)$.
So whether $x \in A$ or $x \in B \cap C$, $x \in (A \cup B) \cap (A \cup C)$ so $A \subseteq (A \cup B) \cap (A \cup C)$.
....
Let $y \in (A \cup B) \cap (A \cup C)$.
Either $y \in A$ or $y \not \in A$.
If $y \in A$ then $y \in A \cup (A \cap B)$.
If $y \not \in A$...
then $y \in A \cup B$ so $y \in A$ OR $y \in B$. If $y \not \in A$ then $y \in B$.
likewise $y \in A \cup C$.  If $y \not \in A$ then $y \in C$so $y \in B$ and $y \in C$.
So $y \in B \cap C$.  So $y \in A \cup (B \cap C)$.
So no matter how $y \in (A\cup B) \cap (A\cup C)$ it follows $y \in A \cup (B \cap C)$
So $(A \cup B) \cap (A\cup C) \subseteq  A \cup (B \cap C)$
And $A \cup (B \cap C) \subseteq (A \cup B) \cap (A\cup C)$
So $A \cup (B \cap C) = (A \cup B) \cap (A\cup C)$
