Proving $A \subset B \implies A \cup B = B$ I have to prove this:
$A \subset B \implies A \cup B = B$
One direction:
$x\in A \subset B$
$\implies x \in A \land x \in B$
$\implies x \in A \cup B$ 
$\implies x \in (A \cup B = B)$
Other direction:
$x \in (A \cup B = B)$
$\implies x \in A \land x \in B$  (I used $\land$ here instead of $\lor$ because it equals B, so it has to be in both sets.)
$\implies x \in A \subset B$ 
Is my proof correct or are there some steps missing.
If not I hope you could show me the right direction.
 A: Sorry, but it's incorrect.
You want to show that “$A\subset B$” if and only if $A\cup B=B$”.
In order to do this, you have to prove


*

*If $A\subset B$, then $A\cup B=B$

*If $A\cup B=B$, then $A\subset B$
Proof of 1
Suppose $A\subset B$. 


*

*Take $x\in A\cup B$; then $x\in A$ or $x\in B$; if $x\in A$, then, by assumption, $x\in B$. In either case, $x\in B$. Hence $A\cup B\subset B$

*Take $x\in B$; then $x\in A\cup B$. Hence $B\subset A\cup B$.


The two points above show that $A\cup B=B$.
Proof of 2
Suppose $A\cup B=B$. Take $x\in A$; then $x\in A\cup B$, so $x\in B$. Therefore $A\subset B$.
A: What you should have done for the proofs:
Proof of 1
Suppose $A\subset B.$
Take $x\in A\cup B.$ Then we have two possibilities: $x\in A$ or $x\in B.$ If $x\in A,$ $x\in B$ by assumption, and the case when $x\in B$ immediately holds. So $A\cup B\subset B.$ If $x\in B,$ then $x\in A\cup B,$ and so $B \subset A\cup B.$  Thus $A\cup B=B.$ 
Proof of 2
Suppose $A\cup B=B.$
Let $x\in A$. Then $x\in A\cup B.$ By hypothesis, $x\in B.$ Thus $A\subset B.$
A: ($\implies$) Suppose that $ A\subseteq B $.
Take any $ x\in A \cup B $, so $ x\in A $ or $ x\in B $. Since $ A\subseteq B $ and $ x\in A $, so $ x\in B $. Then, we have $ A\cup B \subseteq B $.
Take any $ x\in B $. Since $ A\subseteq B $ and $x\in B$, so $x\in A$ or $x\in B$. Then, we have $B\subseteq A\cup B$.
Since $ A\cup B \subseteq B $ and $B\subseteq A\cup B$, so $A\cup B=B$.
($\impliedby$) Suppose that $A\cup B= B$.
Take any $ x\in A\cup B $, so $x\in A$ or $x\in B$, in particular, $x\in A$. Since $A\cup B=B$ and $x\in A\cup B$, then $x\in B$.
Since for any $x\in A \implies x\in B$, then we have $A\subseteq B$
A: Below a proof using the natural deduction method. 
The goal is read as a simple implication : if A is included in B , then ... ( not as a double implication). 
A: Thanks for the help. I hope this is better.
Let $x \in A \subset B$.
$\Rightarrow x \in A \land x \in B \Rightarrow x \in (A \cup B) \Rightarrow x \in B \Rightarrow x \in (A \cup B = B)$. 
Assume $A \cup B = B$. 
Let $x \in A \Rightarrow x \in (A \cup B) \Rightarrow x \in B \Rightarrow x \in (A \subset B)$.
Could you please show me any mistakes or ways to improve it.
