Is my proof of $-(-a)=a$ correct? I'm trying to prove theorem $1.4$ with the following:



$$a=a$$
$$a-a=\stackrel{0}{\overline{a-a}}\tag{Ax.5}$$
$$a-a=0\tag{Inverse def.}$$
$$a+(-a)=0\tag{Thm 1.3}$$
Here I thought about packing the $a$ with a minus sign and then it would yield a new minus sign on Its inverse due to the inverse definition.
$$(-a)+(-(-a))=0$$
Adding $a$ to both sides.
$$\stackrel{0}{\overline{a+(-a)}}+(-(-a))=a$$
$$-(-a)=a$$
Is my proof correct?

Better wording:
By theorem 1.2.  For every a there is a unique -a such that a + (-a) = 0.  Likewise for (-a) there is a unique -(-a) such that (-a) + (-(-a)) = 0.
Therefore:
(-a) + (-(-a)) = 0
a + (-a) + (-(-a)) = a+ 0 = a
By associativity:
[a + (-a)] + (-(-a))  = a so
0 + (-(-a)) = (-(-a)) = a.
