By left distributivity and the role of the zero element we can write

a * 0 + a * 0 = a * (0 + 0) = a * 0 = a * 0 + 0,

so that

a * 0 + a * 0 = a * 0 + 0.

The cancellation law for groups allows us to conclude:

a * 0 = 0, as required.

Similarly one shows 0 * a = 0.

By Part 1,

a * (b + -b) = a * 0 = 0.

Using distributivity to expand the left-hand side, we find:

a * b + a * (-b) = 0,

from which we derive -(a * b) = a * (-b).

The other equalities follow in a similar way.

According to Part 2,

(-a) * (-b) = -((-a) * b),

so (-a) * (-b) is the inverse of (-a) * b.

But from Part 2 we also conclude that a * b is the inverse of (-a) * b. Since (additive) inverses are unique, we are done.

a + ((-1) * a) = 1 * a + ((-1) * a) = (1 + -1) * a = 0 * a = 0.

So (-1) * a is the additive inverse of a, i.e., -a = (-1) * a.