How to prove that $A⊆B$ means that $A∪B=B$ How does one prove that $A⊆B$ means that $A∪B=B$ ? I can understand it in my head but I don't know how you'd put down in logic notation.
 A: Since $B \subseteq A\cup B$ it suffices to show the other containment. 
Let $x \in A \cup B$. Then by definition, this means either $x \in A$ or $x \in B$. The former, by assumption, implies $x \in B$. Therefore, $x \in A \cup B$ implies $x \in B$. 
A: $A \cup B = B$ is equivalent to the following statement
$$ A \cup B \subseteq B \;\;\; \text{and} \;\;\;  B \subseteq A \cup B. $$

The first one, $A \cup B \subseteq B$, means that every element $x \in A \cup B $ must also be in $B$. If $x \in A \cup B $, that either $x \in A $ or $x \in B $. Finally, notice that, if $x \in A$, since $A \subseteq B$, by assumption, it follows that $x \in B$.

The second one, $B \subseteq A \cup B$, is trivial. 
A: Let's do it formally.
\begin{align*}
A\cup B = B 
  &\Leftrightarrow \forall a\colon((a\in B \lor a\in A)\Leftrightarrow a\in B)
&&\text{Definition}
\\&\Leftrightarrow \forall a\colon(((a\in B\lor a\in A)\Rightarrow a\in B)\\&\quad\quad\land((a\in B\lor a\in A)\Leftarrow a\in B)
&&\text{Tautology in And}
\\&\Leftrightarrow \forall a\colon(((a\in B\lor a\in A)\Rightarrow a\in B)
&&\text{Def. of Implication}
\\&\Leftrightarrow \forall a\colon\neg(((a\in B\lor a\in A)\land a\notin B)
&&\text{Distributivity}
\\&\Leftrightarrow \forall a\colon\neg( ((a\in B)\land(a\notin B))\lor ((a\in A)\land (a\notin B)))
&&\text{Contradiction in Or}
\\&\Leftrightarrow \forall a\colon\neg( (a\in A)\land (a\notin B))
&&\text{Def. of Implication}
\\&\Leftrightarrow \forall a\colon (a\in A \Rightarrow a\in B)
&&\text{Definition}
\\&\Leftrightarrow A\subseteq B
\end{align*}
