Prove that $(a^{-1})^{-1}=a$ How does one prove that if $a \ne 0$, then $(a^{-1})^{-1}=a$?
My friend (I'm trying to help her) has in her class notes:

$a+(-a)=0$ and $(-a)+a=0$ implies that $a=-(-a)$ by the uniqueness
  theorem.

Why does $a+(-a)=0$ and $(-a)+a=0$ imply $a=-(-a)$? How do I use that to prove $(a^{-1})^{-1}=a$?
I started out by writing $ab=1$ implies that $b=a^{-1}$. Then, mimicking the teacher's notes I write $a a^{-1}=1$ and $a^{-1} a=1$. I guess next I could just write the conclusion, but I don't understand the reason why it's so. Thanks.
 A: By definition, $-x$ is the unique number that, when added to $x$, gives $0$. 
So what is $-(-x)$? By definition, it is the unique number $y$ such that
$$(-x)+y = y+(-x) = 0.$$
Now, as it happens, setting $y=x$ works! That means that $-(-x)$ has to be $x$, because $x$ works, and there is one and only one thing that works.
The same thing happens with $a^{-1}$. By definition, $a^{-1}$ is the unique element that, when multiplied by $a$, gives $1$. What is $(a^{-1})^{-1}$? By definition, it is the unique number $b$ such that
$$(a^{-1})b = b(a^{-1}) = 1.$$
Now, as it happens, setting $b=a$ works! That means that $(a^{-1})^{-1}$ is $a$, because $a$ "works", and there is one and only one thing that works.
(How do we know there is one and only one thing that works? Suppose that $x+y=y+x=0$ and $x+z=z+x=0$; then
$$y = y+0 = y+(x+z) = (y+x)+z = 0+z = z,$$
so $y=z$. Similarly, suppose that $xy = yx = 1$ and $xz=zx = 1$; then
$$y = y1 = y(xz) = (yx)z = 1z = z$$
so $y=z$; that is, there is at most one thing that works; since there is at least one thing that works, there is one and only one thing that works)
