For any $x \in R$ $$x\cdot 0 = 0\cdot x =0$$ Proof: $$x\cdot 0=x\cdot (0+0)=x\cdot 0+x\cdot 0 \implies x\cdot 0=x\cdot 0+(-(x\cdot 0))=0$$
▸ From here, by the way, it can be observed that if $x\in R\setminus 0$, then $x^{-1}\in R\setminus 0$
How can the statement in the first line, along with the subsequent proof, lead to the statement in ▸?
By definition, $x\cdot x^{-1}=1\neq 0$. Thus if $x\cdot 0=0$, then $x^{-1}\neq 0$