Wondering if proof is proper so I have been working on learning some new math in order to prepare for next year. I have been trying to learn proofs, and doing practice questions however the only problem is there are not answers. I have tried some of the questions, and am wondering if my answers are acceptable, and if not, what I need to work on? thanks


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*If A and B are sets, show that $ A\subseteq B $ if and only if A $ \cap $ B= A 


My proposed proof:
Let x be an element of A, then x is also an element of B ( due to the fact that A is  subset of B). Thus A is an element of A $\cap$B. If there was an element 'd' that was in A but not in B then 'd' would not be in $A \cap B $, same for an element in B that is not in A. Thus A $\cap B $ $\subseteq$A. And because no element of A is not in A $\cap $ B because A is a subset of B, it follows that A $\subseteq$$A\cap$B. 
Hence A=$A\cap$B.
QED


*

*Prove the second de Morgan Law
A \ (B $\cap$ C)= (A \ B) $ \cup$ (A \ C)


my proposed proof:
let x be an element of A \ (B $\cap$ C). Then x is an element of A but not an element of(B $\cap$ C). Hence x is in A and is in most of one of B or C. x can be in A and in B, but not C or x can be in A and in C but not B or lastly x can be in A and not in B and not in C.
in the first case x is in A but not C hence x is in A \ C
in the second x in A but not B hence in A \ B
and the last case x is in A \ B and in A \ C
In any of the cases, x is contained in (A \ C) $\cup$ (A \ B)
conversely if x is in (A \ C) $\cup$ (A \ B) x can be in ( A \ C), (A \ B) or both. either of the ways is x contained in the union of (A/B) and (A/C)
Thanks everyone for there help and suggestions.
 A: For the first proof: It is a little unclear what is being proved. This may seem easier to follow. Note the style of explicitly breaking up the proof into two separate implications. This usually adds much to the clarity of the proof.
Proof: ($\Rightarrow$) Suppose $A\subset B$; we must show $A\cap B=A$. It is always true that $A\cap B\subset A$, so it suffices to show $A\subset A\cap B$. Let $a\in A$. By hypothesis, $A\subset B$, so $a\in B$ as required.
($\Leftarrow$) Suppose $A\cap B=A$; we must show $A\subset B$. Let $a\in A$. By hypothesis, we have $a\in A\cap B$. It is always true that $A\cap B \subset B$, so $a\in B$ as required. $\blacksquare$
A: I will try to simplify your first proof a little.


*

*If A and B are sets, show that $ A\subseteq B $ if and only if A $ \cap $ B= A 


My proposed proof:
Suppose $A\subseteq B$. Let $x$ be an element of A. Then $x$ is also an element of B. Thus $x$ is an element of $A \cap B$. So $A\subseteq A\cap B$.
If there was an element '$d$' that was in $A$ but not in $B$ then '$d$' would not be in $A \cap B $ . (Note: here you are using a sort of contrapositive argument). Thus $A \cap B \subseteq A$. 
Hence $A=A\cap B$.
Now suppose $A=A\cap B$.  Let $a\in A$.  Then $a\in B$. So $A\subseteq B$.
QED
