Show that these three statements are logically equivalent. show that these three statements are logically equivalent.
$A \subseteq B, A \cup B = B, A \cap B = A$.
I am unsure how to begin this, so i have set up as follows
First I must show that.
$A \subseteq B \implies A \cup B = B  $
since $A$ is a subset of $B$, every element of $A$ is an element of $B$. We have
$A \cup B = \{x \in A \text{ or }x \in B\} = B$.
I have completed further, but i have a feeling that i have not proven anything so far. Could someone be of assistance please. 
 A: Your first proof should be fine, although you might want to flush it out a little more and argue for equality using double containment. Since $A\subset B$, it is impossible that $A\cup B$ contains an element outside of $B$, which is to say $A \cup B \subset B$. On the other hand, it is obvious that $B \subset A\cup B$. By double containment you conclude that $A\cup B = B$. The next things you need to show is that 

If $A \cup B = B$ then $A\cap B = A$.

You can make a similar argument and use double containment to prove that $A\cap B = A$. Lastly, you need to show that 

If $A\cap B = A$ then $A\subset B$. 

Once you have done this, you will have shown that all three statements are logically equivalent. 
A: First: $A \subset B \Rightarrow A\cup B = B$
Take $x \in A \cup B$. Then because $A \subset B$ follows $x \in B$. So $A \cup B \subset B$.
Now take $x \in B$. It follows directly from the definition that $x \in A \cup B$. Now also $ B \subset A \cup B$. Hence, $A \cup B = B$
Second: $A \cup B = B \Rightarrow A \cap B = A$
Take $x \in A \cap B$. Then it's clear that $x \in A$. So $A \subset A\cap B$
Now take $x \in A$. Then $x \in A \cup B$, so from the assumed statement follows $x \in B$. So $A \cap B \subset A$
Hence, $A \cap B = A$
Last: $A \cap B = A \Rightarrow A \subset B$
Take $x \in A$. Then $x \in A \cap B$, so $x \in B$.
Hence, $A \subset B$.  
That's it. Hope you understood it, and if not I would be pleased to help ;)
