Using field and order axioms prove ab ab Using field and or

Solution

proof is :

First we prove that 0.a=0. Indeed by distributive axiom we have

0.a=(0+0).a=0.a+0.a,

then by adding the additive inverse of 0.a to both sides we find 0.a=0.

Now we have (1).a=-a, in fact:

0=(1+(1)).a=a+(1).a,

so

(1).a=a.

Finally we prove your equality:

(a)+(b)=(1).a+(1).b=(1).(a+b)=(a+b).

That\'s all.