I'm supposed to prove this. The answer key doesn't tell me anything more than to add zero:
$(a+(-a))$
to:
$(-1)\cdot a$
and use that:
$a=a\cdot 1$
I have these axioms to my disposal:
(A1) for all $x,y$ we have $+y=y+x$
(A2) for all $x,y,z$ we have $(x+y)+z=x+(x+z)$
(A3) there exists a number $0$ so that for all $x$ we have $x+0=x$
(A4) for all $x$ there exists a number $a$ so that $x+a=0$
(M1) for all $x,y$we have $x*y=y*x$
(M2) for all $x,y,z$ we have $(x*y)*z=x*(y*z)$
(M3) there exists a number $1=/=0$ so that for all $x$ we have $x(1/x)=1$
(M4) for all $x=/=0$ there exists a number $b$ so that $x(1/x)=1$
(AM) for all $x,y,z$ we have $z*(x+y)=z*x+z*y)$
$a ∈ R$