\item
Let $S, T, U$ and $V$ be sets. Prove that if $S\subseteq T$ and $T \subseteq U$ and $V \subseteq U$, then $S\cup V \subseteq U$.
\begin{proof}
	Let $S, T, U$ and $V$ be sets such that $S\subseteq T$ and $T \subseteq U$ and $V \subseteq U$.
	If $x \in S\cup V$ then $x \in S$ or $x \in V$. \\
	Case: $x \in S$ \\
	Let $x \in S$. Since $S \subseteq T$ then $x \in T$. Since $x \in T$ and $T \subseteq U$ then $x \in U$. \\
	Case: $x \in V$ \\
	Let $x \in V$. Since $V \subseteq U$ then $x \in U$. \\
	In both cases $x \in U$. \\
	$\therefore$ if $S, T, U$ and $V$ be sets such that $S\subseteq T$ and $T \subseteq U$ and $V \subseteq U$, then $S\cup V \subseteq U$.
\end{proof}
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\item
Let $A$ and $B$ be sets. Prove that $(A \cap B) \cup (A - B) = A$.
\begin{proof}
	Let $A$ and $B$ be sets.
	Prove $A \subseteq (A \cap B) \cup (A - B)$: \\
	Let $x \in A$. There are two cases either $x \in B$ or $x \notin B$. If $x \in B$, then $x$ is in both $A$ and $B$, so $x \in A \cap B$. \\
	If $x \notin B$, then $x \in A$ and $x \notin B$, so $x \in A-B$. \\
	In all cases, $A \subseteq (A \cap B) \cup (A - B)$. \\
	Prove $ A \supseteq (A \cap B) \cup (A - B)$:
\end{proof}
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