 Solution

Take $(x,y)\in E_r(K)\setminus \{O\}$. Because $v(x)<0$ and $p,q$ have degrees $d,d-1$, their leading terms dominate, givingThe polynomial $p^{\prime }q-p{q}^{\prime }$ has degree $2d-2$ with nonzero leading coefficient, the same degree as $q^2$. Their quotient consequently has valuation zero at $x$, soThe isogeny sends $O$ to $O$, hence $\varphi (E_r(K))\subseteq E_r^{\prime }(K)$.