How can I prove if $A\subseteq B$, then $A\cup B=B$? I need to prove that  $$A\subseteq B \implies A\cup B=B$$ I defined the subset relation as the statement $x\in A\Rightarrow x\in B$. I tried to convert the claim into a logic statement, then proceeded to simplify the statement $$(x\in A\Rightarrow x\in B)\Rightarrow(x\in A\vee x\in B),$$ and I simplified it to $x\in A\vee x\in B$. Only after I did this did I realize that this didn't really prove anything.
This fact seems to be taken axiomatically—I haven't been able to find any proofs for it.
I found that it is proposed under Wikipedia's Proposition 8 under ‘Algebra of Sets’ that this is true, and equivalent to the subset relation. A proof of this, however, escapes me.
I also realize that it would be sufficient to be able to convert the antecedent to the consequent or vice versa (which would result in the tautology $p\Rightarrow p$ and thus prove the statement), though I do not know how to do this.
 A: Notice that 
$$B \subseteq B \implies B \subseteq B \cup A = A \cup B \implies B \subseteq A \cup B$$
and $$ A \subseteq B \implies A \cup B \subseteq B\cup B = B \implies A \cup B \subseteq B $$
A: If you really want to do an element-chasing proof, then this is how you might go about it:
Start by supposing that $A\subseteq B$ (i.e., $x\in A\to x\in B$). 
Suppose $x\in A\cup B$. Then $x\in A$ or $x\in B$. If $x\in A$, then $x\in B$ because $A\subseteq B$. If $x\in B$, then obviously $x\in B$. In either case, we must conclude $x\in A\cup B\to x\in B$. Thus, $A\cup B\subseteq B$. 
Suppose $x\in B$. Then we clearly must have $x\in A\cup B$. That is, $x\in B\to x\in A\cup B$. Thus, $B\subseteq A\cup B$.
By mutual subset inclusion, we have that $A\subseteq B\to A\cup B=B$. $\blacksquare$
A: A proof by contradiction.  
Given $A \subseteq B $ assume $A \cup B \not= B$.  Then there is $x \in A\cup B$ and $x \notin B$ or $x \in B$ and $x \notin A \cup B$.  Clearly the later contradicts the definition of set union.  
If its the former then $x \in A\cup B \implies x \in A$ since $x \notin B$. But $A \subseteq B \implies x \in B$ which makes the former impossible by contradiction as well. 
Since there is no $x$ in $A\cup B$ that is not in $B$ and vice versa we must conclude that $A \cup B = B$.    
