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?

Your proof is correct, but it looks like you're trying to be as precise as possible, so I'll write out a proof you might find helpful. Let $a$ be a real number. By axiom $5$, we there exists a real number $b$, satisfying $a+b=0$.  
Since $b$ is a real number, it also has an inverse, which we may call $c$. The aim is to show $c=a$. Since $a+b=0$, we have $$(a+b)+c=0+c=c$$
And then by axiom $2$ (associativity) we deduce that $$a+(b+c)=c$$
But since $c$ was the inverse of $b$, $b+c=0$, so $a+0=c \implies a=c$. Hopefully that's helpful.
