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?

What is "packing with a minus sign"? This step seems a little obscure. You are acting like there is an operation which is "putting minus in front", even though (in this stage of axioms) $-a$ is a symbol in itself.
For proving $-(-a)=a$, we have to prove (by definition) that $-a+a=0$. But this is trivial due to commutativity and the characterization of $-a$.
Also, more leisurely:
Proof: We have that $a+(-a)=0$. Now look again. $\blacksquare$
