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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)
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