At the point $(0,0)$ of $w'^2=-s(s^m+1)(\bar\alpha s^m+\alpha)$, the valuation satisfies $v(s)=v(w')^2$

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At the point $(0,0)$ of $w'^2=-s(s^m+1)(\bar\alpha s^m+\alpha)$, the valuation satisfies $v(s)=v(w')^2$ · Prove2Me