I tried to prove that:
$ \forall a \in \mathbb{R}: (-(-a)) = a $
using the following field axioms:
(1) Commutative law (addition)
(2) Associative law (addition)
(3) Existence and uniqueness of the number zero (addition)
(4) Commutative law (multiplication)
(5) Associative law (multiplication)
(6) Distributive law
(7) Existence and uniqueness of the number 1 (multiplication)
I know that,
$ \forall a \in \mathbb{R}: a + (-a) = 0 $
The proof:
Let $a \in \mathbb{R}$.
$(-(-a))$
$= (-1)(-1)a $. using (6) and (7)
$= (-1+1-1)(-1+1-1)a $. using (3)
$= ((-1)(-1)+(-1)+(-1)(-1)+(-1)+1+(-1)+(-1)(-1)+(-1)+(-1)(-1))a $. using (6)
$= ((-(-1))+(-1)+(-(-1))+(-1)+(-(-1))+(-1)+(-(-1))+(-1)+1)a$. using (6) and (1)
$= ((-1)+(-(-1))+(-1)+(-(-1))+(-1)+(-(-1))+(-1)+(-(-1))+1)a$. using (1)
$= (4(-1)+4(-(-1))+1)a$. using (6)
$= (4(-1)-4(-1)+1)a$. using (2)
$= (0 + 1)a$
$= a$. using (6) and (7)
Is the proof correct? I'm almost sure there is a simpler way using the third axiom.