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)