# Equivalent conditions for commutativity in rings with 1

I have a problem with understanding the proof of theorem $3.1$. Could anyone help me with this?

$(P_1)$ $y^{s}[x,\, y]=\pm x^{p}[x^{m},y^{n}]^{r}y^{q}$ where $m>1,r>0,n\geq0, s\geq0, p\geq0 , q\geq0$

I don't understand why the equation $y^{s}[x^{m},y]-x^{(m-1)(p+m-1)}mx{}^{m-1}(\pm x^{p}[x^{m},y^{n}]y^{q})$ is equal to this equation $=y^{s}[x^{m},y]\mp x^{mp}[x^{m^{2}},y^{n}]y^{q}=0$ and why this is equal to zero.

From the lemma $3.2$ I know that $mx{}^{m-1}[x^{m},y^{n}]=[x^{m^{2}},y^{n}]$ and after calculations there still left $x^{mp}x^{(m-1)^2}$. I don't know why $x^{(m-1)^2}$ disappeared.

Thank you.

-
Could you at least repeat the assumptions in the question. And possibly also the notations? – Tobias Kildetoft Aug 10 '13 at 18:09
@ Tobias Kildetoft: I did it. – Monika Aug 10 '13 at 19:09
Must the text be such an illegibly small graphics? – Hagen von Eitzen Aug 11 '13 at 18:00
@Hagen von Eitzen: Sorry, it should be better from now. – Monika Aug 11 '13 at 21:18