[proofplan] The proof uses the explicit inverse formula in the group law on a short Weierstrass elliptic curve. For an affine point $(x,y)$, the inverse is $(x,-y)$. Since the given point has $y$-coordinate $0$, it is equal to its own inverse, and the group inverse axiom then gives $P+P=O$. [/proofplan] [step:Identify the inverse of the point with zero $y$-coordinate] For the group law on the short Weierstrass curve $E:y^2=x^3+ax+b$, the inverse of an affine point $Q=(x,y)\in E(k)$ is \begin{align*} -Q=(x,-y). \end{align*} Applying this to $P=(x_1,0)$ gives \begin{align*} -P=(x_1,-0). \end{align*} Since $0$ is the additive identity of the field $k$, we have $-0=0$, and therefore \begin{align*} -P=(x_1,0)=P. \end{align*} [guided] The group law on a short Weierstrass elliptic curve has a simple formula for inverses: if $Q=(x,y)$ is an affine point of $E(k)$, then \begin{align*} -Q=(x,-y). \end{align*} Geometrically, this is reflection across the $x$-axis; algebraically, it preserves the curve equation because \begin{align*} (-y)^2=y^2=x^3+ax+b. \end{align*} Thus $(x,-y)$ is again a point of $E(k)$. Now take the specific point $P=(x_1,0)$. Substituting $x=x_1$ and $y=0$ into the inverse formula gives \begin{align*} -P=(x_1,-0). \end{align*} The element $0\in k$ satisfies $-0=0$, so \begin{align*} -P=(x_1,0)=P. \end{align*} Thus $P$ is its own inverse in the elliptic curve group $E(k)$. [/guided] [/step] [step:Use the inverse axiom in the elliptic curve group] The set $E(k)$ with the elliptic curve addition law is an abelian group with identity element $O$. Since $-P=P$, the inverse axiom gives \begin{align*} P+P=P+(-P)=O. \end{align*} Hence $P$ has order dividing $2$, so $P\in E(k)[2]$. [/step]