invertible elements in a Ring I'm reading a proof in the book Basic Abstract Algebra (by P. B Bhattacharya, S. R. Nagpaul, S. K. Jain page 172) and I'm trying to understand the proof from a  result. The result states:
let x be a nonzero element of a ring R with 1. if there exist a unique element $y \in R$  such that xyx=x, then xy=1=yx; that is,x is invertible in R.
Solution: 
let $xr=0$, $r \in R$. Then $x(y+r)x=x$ and hence y+r=y. Thus, $xr=0$ implies $r=0$. Now $xyx=x$  gives $x(yx-1)=0$. Thus, $yx-1=0$, which completes the solution
My question is why do they have to prove $xr=0 \Rightarrow r=0$ in the solution? 

Knowing that $x$ isn't a left zero divisor provides a lever to show that the one-sided inverse is two-sided. 
To be honest, I think the whole strategy is difficult to anticipate before you have seen the trick. The best justification I can come up with is it's like learning to walk before you run: "if you want to prove it's left invertible and can't, can you at least show it's not a left zero divisor? " That is necessary for it to be left invertible, so it would be a step in the right direction. 
Then after it turns out to be true: "Oh look, now I can show it is left invertible using that fact with $x(yx-1)=xyx-x=0$..."
