Mathematical Proof/Appendix/Answer Key/Mathematical Proof/Methods of Proof/Constructive Proof

Problem 1.1) First, we wish to show that $A\cup (B\cap C)\subset (A\cup B)\cap (A\cup C)$. Let $x\in A\cup (B\cap C)$. Then $x\in A$ or $x\in B\cap C$.

case 1: $x\in A$
$x\in A\subset A\cup B$ so that $x\in A\cup B$
$x\in A\subset A\cup C$ so that $x\in A\cup C$
$x\in A\cup B$ and $x\in A\cup C$ so that $x\in (A\cup B)\cap (A\cup C)$

case 2: $x\in B\cap C$
$x\in B$ and $x\in C$
$x\in B\subset A\cup B$so that $x\in A\cup B$
$x\in C\subset A\cup C$ so that $x\in A\cup C$
$x\in A\cup B$ and $x\in A\cup C$ so that $x\in (A\cup B)\cap (A\cup C)$

Since in both cases, $x\in (A\cup B)\cap (A\cup C)$, we know that $A\cup (B\cap C)\subset (A\cup B)\cap (A\cup C)$

Now we wish to show that $(A\cup B)\cap (A\cup C)\subset A\cup (B\cap C)$. Let $x\in (A\cup B)\cap (A\cup C)$. Then $x\in A\cup B$ and $x\in A\cup C$.

case 1a: $x\in A$
$x\in A\subset A\cup (B\cap C)$ so that $x\in A\cup (B\cap C)$

case 1b: $x\in B$
We can't actually say anything we want to with just this, so we have to also add in $x\in A\cup C$
case 2a: $x\in A$ : SEE CASE 1a
case 2b: $x\in C$
We now have $x\in B$ and $x\in C$ so that $x\in B\cap C$
Of course, since $x\in B\cap C\subset A\cup (B\cap C)$, it follows that $x\in A\cup (B\cap C)$.
Since both cases 2a and 2b yield $x\in A\cup (B\cap C)$, we know that it follows from 1b.

Since in both cases 1a and 1b, $x\in A\cup (B\cap C)$, we know that $(A\cup B)\cap (A\cup C)\subset A\cup (B\cap C)$.

Since both $A\cup (B\cap C)\subset (A\cup B)\cap (A\cup C)$ and $(A\cup B)\cap (A\cup C)\subset A\cup (B\cap C)$, it follows (finally) that $A\cup (B\cap C)=(A\cup B)\cap (A\cup C)$.

--will continue later, feel free to refine it if you feel it can be--