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Proved using field axioms of real numbers
take y = -x
from field axiom 4, x + 0 = x
-> 0 = x - x (eq 1)
-> 0 = x + (-x)
-> 0 = x + y (eq 2)
-x is short form of 0-x
Now -(-x) = 0-(-x)
= 0-(0-x) = x+y -(x+y-x) from eq 2 = x+y-(y+x-x) from field axiom 1 = x+y-(y+(x-x)) from field axiom 2 = x+y-(y + 0) using eq 1 = x+y-(y) from field axiom 4 = x+y-y = x+(y-y) from field axiom 2 = x + 0 (since y-y = 0 as a consequence of field axiom 4 as shown in eq 1, for any real y) = x (from field axiom 4)