Minus times minus results in a plus, The reason for this, we needn't discuss.
  - Ogden Nash 

Let a and b be in a field (the same proof is possible in a ring, but due to the possible lack of commutativity of multiplication in a ring, it takes more steps)

Lemma 1: a*0 = 0*a = 0
 a*0 = a*(0 + 0) <0 = 0+0>
 a*0 = a*0 + a*0 <distributivity in a field>
 0 = a*0

Lemma 2: -(-(a)) = a
 By definition of additive inverse, -(-(a)) + -(a) = 0.
 But, we know that -(a) + a = 0, so -(-(a)) = a, because additive inverses are unique.
 
Lemma 3: (-a)*(b) = (a)*(-b) = -(a*b)
 0 = a*0 = a*(b+(-b)) = a*b + a*(-b)
 this implies that a*(-b) is an additive inverse of a*b, so a*(-b) = -(a*b)
 Similarly, 0 = 0*b = (a+(-a)*b = a*b + (-a)*b, so (-a)*b = -(a*b)
 so (-a)*(b) = (a)*(-b) = -(a*b)
 
 Prop: (-a)*(-b) = (a*b)
 (-a)*(-b) = -((-a)*b) <by lemma 3.1>
 -((-a)*b) = -(-(a*b)) <by lemma 3.2>
 -(-(a*b)) = a*b <by lemma 2>