Proving "if and only if" statements I have the following proposition: 

$m - n = p - q$ if and only if $m + q = n + p$.

From what I understand, $A = B$ if and only if $C = D$ means two statements: 


*

*if $A = B$, then $C = D$ and 

*if $C = D$, then $A = B$. 


So basically, I need to prove the two statements, right? I am learning proofs and will get to study the "if and only if". However, what is its usefulness?
 A: As you said, the sentence "$m−n=p−q$ if and only if $m+q=n+p$" means that:


*

*You can conclude $m−n=p−q$ from $m+q=n+p$

*You can conclude $m+q=n+p$ from $m−n=p−q$


So when we define $$\begin{align}A: m−n=p−q \\B:m−n=p−q\end{align}$$ you have $A \iff B$. The advantage is, that now $A$ and $B$ share the same truth value (you call "$A$ is equivalent to $B$").
Some benefits from the equivalence of statements:


*

*Imagine you want to define a property like continuity of a function. If $A(x)$ is the definition of continuity (i.e. $A(f)$ is true whenever $f$ is continuous) and you have another proposition which $\forall x(A(x) \iff B(x))$ then you can also choose $B(x)$ as the definition of continuity

*Imagine you want to prove $a^2 + b^2 \ge 2ab$. You can write $$a^2 + b^2 \ge 2ab \iff a^2 -2ab + b^2 \ge 0 \iff (a+b)^2 \ge 0$$ Because you can easily prove that the last inequality is true and you only used $\iff$ also the first inequality must be true...

A: You are trying to prove an if and only if, iff, biconditional, etc (whatever other terminology you want to use) statement. In this case, you can cut off some of the work you have to do for yourself by noting that equality (i.e. $=$) goes both ways. That is, $2_1 = 2_2$ really means $(2_1\to 2_2) \leftrightarrow (2_2\to 2_1)$. With this in mind, let's actually take a look at your problem specifically. 
Denote your statements as follows:
$\Omega : m-n=p-q\tag{1}$
and
$\Phi : m+q=n+p\tag{2}$
Now rearrange $(1)$ and $(2)$ as $m-n-p+q=0$ and $m+q-n-p=0$, respectively. Then we may see that the biconditional claim we are making is ultimately that
$$
-(n+p-m-q) = n+p-m-q.\tag{3}
$$
Thus, your claim is true precisely when both sides of $(3)$ equal $0$. This should get you started in your proof (only a very small way to go until you can complete it). 
