Vtyjkyuk

In W. Rudin's Principles of Analysis, 1.14, the proof for $-(-x) = x$ from the basic axioms of a field is hinted at but not shown. Based on the hint I came up with the following:
beginarrayl c
x = x & \
x = x + 0 & additive identity\
x = x + (-x) - (-x) & additive inverse\
x = -(-x)
endarray
Is it OK to do what I did above, use $(-x) - (-x)=0$ by definition of additive inverse? I feel I need an extra step since I used $-x$ for each $x$ in the definition $x + (-x) = 0$. Or maybe my whole approach is wrong?

Your approach is perfectly valid.

$$x = x$$

$$x = x + 0$$
$$x = x + ((-x) - (-x))$$
Associative property of addition

$$x =( x + (-x)) - (-x)$$
$$x = 0 -(-x)=-(-x)$$

Your approach is fine but you jumped two steps.

You have

$x=x+(-x)-(-x) $ so the next should be:

$x=(x+(-x))-(-x) $ associativity of addition

$x=0-(-x)$ additive inverse

$x=-(-x) $ additive identity.

BTW I think a more direct proof would be to note:

$x+(-x)=0$ and $-(-x)+(-x)=0$ so

$x+(-x)=-(-x)+(-x)$ so

$x+(-x)+x=-(-x)+(-x) +x$ from the axiom that $a=biff a+c=a+c $

Then associativity, inverses, and identity will yield

$x+0=-(-x)+0$ and $x=-(-x) $.