[proofplan] We substitute the exact solution into the multi-step method, expand in Taylor series about $t_n$, and recognise the resulting expression as $\rho(e^x) - x\sigma(e^x)$ evaluated at $x = h$. The local truncation error vanishes to order $p$ precisely when this generating function vanishes to order $p+1$. [/proofplan] [step:Substitute the exact solution and expand in Taylor series] The method $\sum_{\ell=0}^s \rho_\ell y_{n+\ell} = h\sum_{\ell=0}^s \sigma_\ell f(t_{n+\ell}, y_{n+\ell})$ applied to the exact solution gives: \begin{align*} y(t_n + \ell h) &= \sum_{k=0}^\infty \frac{(\ell h)^k}{k!}y^{(k)}(t_n). \end{align*} Therefore $\sum_\ell \rho_\ell y(t_n + \ell h) = \sum_{k=0}^\infty \frac{h^k}{k!}\left(\sum_\ell \rho_\ell \ell^k\right)y^{(k)}(t_n)$. Recognising $\sum_\ell \rho_\ell \ell^k$ as the $k$-th coefficient in the Taylor expansion of $\rho(e^x)$, the left side equals $\rho(e^x)$ evaluated symbolically at $x = h$. Similarly, $h\sum_\ell \sigma_\ell y'(t_n + \ell h) = x\sigma(e^x)$ in the same sense. The local truncation error vanishes to order $p$ precisely when $\rho(e^x) - x\sigma(e^x) = O(x^{p+1})$. [/step]