[proofplan] The log-ODE one-step approximation $\mathcal{L}^{\lfloor\gamma\rfloor}_{s,t}$ is the time-1 flow of an autonomous ODE on $[0,1]$ whose driving vector field $F^{\lfloor\gamma\rfloor}_{s,t}$ is the composition of the truncated log-signature of $x|_{[s,t]}$ with the Lie-algebra homomorphism $\Phi_f$ associated with $f$. We compare the true RDE flow $y_s \mapsto y_t$ to $\mathcal{L}^{\lfloor\gamma\rfloor}_{s,t}(y_s)$ in three pieces: bound the Lip$^\gamma$ norm of $F^{\lfloor\gamma\rfloor}_{s,t}$ by $\|f\|_{\mathrm{Lip}^\gamma}\|x\|_{p\text{-var};[s,t]}$ using the log-signature bound; apply the bounded-variation Euler local-error bound to compare $\mathcal{L}^{\lfloor\gamma\rfloor}_{s,t}(y_s)$ to the level-$\lfloor\gamma\rfloor$ Euler step $\mathcal{E}^{\lfloor\gamma\rfloor}_{0,1}(y_s; F^{\lfloor\gamma\rfloor}_{s,t}, t)$ on the auxiliary time interval, observing that the two Euler steps coincide algebraically by the tensor-exponential identity $S^{\lfloor\gamma\rfloor} = \exp^\otimes \log^{\lfloor\gamma\rfloor}$; apply the *Local Error of Euler Scheme* theorem to bound $\|y_t - \mathcal{E}^{\lfloor\gamma\rfloor}_{s,t}(y_s; f, x)\|$ directly. The triangle inequality with $C = C_1 + C_2$ closes the argument. [/proofplan]

[step:Bound the Lip$^\gamma$ norm of the log-ODE vector field]Define the log-ODE vector field by composition: for $z \in \mathbb{R}^e$, \begin{align*} F^{\lfloor\gamma\rfloor}_{s,t} : \mathbb{R}^e &\to \mathbb{R}^e, \\ z &\mapsto \Phi_f\!\big(\log^{\lfloor\gamma\rfloor}(S(x)_{s,t})\big)(z), \end{align*} where $\log^{\lfloor\gamma\rfloor}(S(x)_{s,t})$ is the truncated log-signature, lying in the free Lie algebra $\mathfrak{L}^{\lfloor\gamma\rfloor}(\mathbb{R}^d)$, and $\Phi_f$ is the Lie-algebra homomorphism associated with $f$ that takes a Lie polynomial to a vector field on $\mathbb{R}^e$. Standard log-signature estimates give \begin{align*} \big\|\log^{\lfloor\gamma\rfloor}(S(x)_{s,t})\big\|_{\mathfrak{L}^{\lfloor\gamma\rfloor}} \leq C_0(p,\gamma)\,\|x\|_{p\text{-var};[s,t]}, \end{align*} for an explicit constant $C_0(p,\gamma)$. Combining with the Lip$^\gamma$ continuity of $\Phi_f$ in its first argument — i.e. $\|\Phi_f(\xi)\|_{\mathrm{Lip}^\gamma(\mathbb{R}^e)} \leq \|f\|_{\mathrm{Lip}^\gamma}\,\|\xi\|_{\mathfrak{L}^{\lfloor\gamma\rfloor}}$ — we obtain \begin{align*} \|F^{\lfloor\gamma\rfloor}_{s,t}(\,\cdot\,;\, f,\, x)\|_{\mathrm{Lip}^\gamma} \leq C_0(p,\gamma)\,\|f\|_{\mathrm{Lip}^\gamma}\,\|x\|_{p\text{-var};[s,t]}. \end{align*} Absorbing $C_0(p,\gamma)$ into a relabelled constant we record the cleaner form \begin{align*} \|F^{\lfloor\gamma\rfloor}_{s,t}\|_{\mathrm{Lip}^\gamma} \leq \|f\|_{\mathrm{Lip}^\gamma}\,\|x\|_{p\text{-var};[s,t]} \end{align*} (modulo a $p,\gamma$-dependent constant absorbed into the final $C$).[/step]

[guided]The log-ODE one-step approximation $\mathcal{L}^{\lfloor\gamma\rfloor}_{s,t}(y_s; f, x)$ is computed in three substeps: 1. Take the truncated log-signature $\xi := \log^{\lfloor\gamma\rfloor}(S(x)_{s,t})$, an element of the free Lie algebra $\mathfrak{L}^{\lfloor\gamma\rfloor}(\mathbb{R}^d)$. 2. Compose with the Lie-algebra-to-vector-field map $\Phi_f$ to get a vector field $F^{\lfloor\gamma\rfloor}_{s,t} = \Phi_f(\xi)$ on $\mathbb{R}^e$. 3. Solve the autonomous ODE $\dot{u} = F^{\lfloor\gamma\rfloor}_{s,t}(u)$ from $u(0) = y_s$ on the time interval $[0,1]$ and read off $u(1)$. To control errors, the first job is to control the regularity of the vector field $F^{\lfloor\gamma\rfloor}_{s,t}$, so we know what we are integrating. The log-signature bound (a standard rough-path estimate, sometimes called the **factorial decay of the log-signature** or **Chen's lemma**) gives \begin{align*} \big\|\log^{\lfloor\gamma\rfloor}(S(x)_{s,t})\big\|_{\mathfrak{L}^{\lfloor\gamma\rfloor}} \leq C_0(p,\gamma)\,\|x\|_{p\text{-var};[s,t]}, \end{align*} where the constant $C_0$ depends on $p$ (the rough-path roughness) and $\gamma$ (the truncation level). The inequality says the log-signature does not grow faster than a single power of the $p$-variation, despite being built from up to $\lfloor\gamma\rfloor$-fold iterated integrals. The map $\Phi_f$ takes a Lie polynomial $\xi \in \mathfrak{L}^{\lfloor\gamma\rfloor}(\mathbb{R}^d)$ to a vector field on $\mathbb{R}^e$. It is a Lie-algebra homomorphism (sending Lie brackets of polynomials to Lie brackets of vector fields), and it is **linear in $\xi$** with the operator-norm bound \begin{align*} \|\Phi_f(\xi)\|_{\mathrm{Lip}^\gamma(\mathbb{R}^e)} \leq \|f\|_{\mathrm{Lip}^\gamma}\,\|\xi\|_{\mathfrak{L}^{\lfloor\gamma\rfloor}}. \end{align*} This bound says $\Phi_f$ has operator norm $\|f\|_{\mathrm{Lip}^\gamma}$ as a linear map from Lie polynomials of degree $\leq \lfloor\gamma\rfloor$ to Lip$^\gamma$ vector fields on $\mathbb{R}^e$. The hypothesis $f \in \mathrm{Lip}^\gamma$ with $\gamma > p \geq 1$ is consumed to give finiteness of $\|f\|_{\mathrm{Lip}^\gamma}$. Composing the two bounds and absorbing $C_0$ into the eventual constant $C$, \begin{align*} \|F^{\lfloor\gamma\rfloor}_{s,t}\|_{\mathrm{Lip}^\gamma} \leq \|f\|_{\mathrm{Lip}^\gamma}\,\|x\|_{p\text{-var};[s,t]}. \end{align*} This is exactly the form of bound we need for the next step, where we will run the level-$\lfloor\gamma\rfloor$ Euler scheme on the auxiliary ODE driven by the elementary bounded-variation path $t$ on $[0,1]$ and apply local-error estimates that are linear in (a power of) the vector-field Lip$^\gamma$ norm.[/guided]

[step:Compare the log-ODE solution to the Euler step on the auxiliary ODE]Consider the auxiliary bounded-variation driving signal $t : [0,1] \to \mathbb{R}$, $r \mapsto r$, with canonical rough lift. Apply the level-$\lfloor\gamma\rfloor$ Euler scheme $\mathcal{E}^{\lfloor\gamma\rfloor}_{0,1}$ to the ODE \begin{align*} \dot{u}(r) = F^{\lfloor\gamma\rfloor}_{s,t}(u(r)), \qquad u(0) = y_s, \end{align*} on $[0,1]$. Since the driving path is the identity on $[0,1]$ (a $1$-variation path), the bounded-variation local-error bound applies: there exists $C_1 = C_1(\gamma)$ such that \begin{align*} \|\mathcal{L}^{\lfloor\gamma\rfloor}_{s,t}(y_s;\, f,\, x) - \mathcal{E}^{\lfloor\gamma\rfloor}_{0,1}(y_s;\, F^{\lfloor\gamma\rfloor}_{s,t},\, t)\| \leq C_1\,\|F^{\lfloor\gamma\rfloor}_{s,t}\|_{\mathrm{Lip}^\gamma}^\gamma\,\|t\|_{1\text{-var};[0,1]}^\gamma. \end{align*} Since $\|t\|_{1\text{-var};[0,1]} = 1$, this simplifies to \begin{align*} \|\mathcal{L}^{\lfloor\gamma\rfloor}_{s,t}(y_s;\, f,\, x) - \mathcal{E}^{\lfloor\gamma\rfloor}_{0,1}(y_s;\, F^{\lfloor\gamma\rfloor}_{s,t},\, t)\| \leq C_1\,\|F^{\lfloor\gamma\rfloor}_{s,t}\|_{\mathrm{Lip}^\gamma}^\gamma. \end{align*} Combining with Step 1, \begin{align*} \|\mathcal{L}^{\lfloor\gamma\rfloor}_{s,t}(y_s;\, f,\, x) - \mathcal{E}^{\lfloor\gamma\rfloor}_{0,1}(y_s;\, F^{\lfloor\gamma\rfloor}_{s,t},\, t)\| \leq C_1\,\big(\|f\|_{\mathrm{Lip}^\gamma}\,\|x\|_{p\text{-var};[s,t]}\big)^\gamma, \end{align*} after relabelling $C_1 = C_1(p, \gamma)$ to absorb the $C_0(p,\gamma)^\gamma$ factor from Step 1.[/step]

[guided]The log-ODE one-step output $\mathcal{L}^{\lfloor\gamma\rfloor}_{s,t}(y_s; f, x)$ is, by definition, $u(1)$ where $u$ solves the autonomous ODE \begin{align*} \dot{u}(r) = F^{\lfloor\gamma\rfloor}_{s,t}(u(r)), \qquad u(0) = y_s, \qquad r \in [0,1], \end{align*} driven by the time-coordinate $r \mapsto r$ on $[0,1]$. We want to compare $u(1)$ to a level-$\lfloor\gamma\rfloor$ Euler step on this auxiliary ODE. The auxiliary driving signal is $t: [0,1] \to \mathbb{R}$, $r \mapsto r$, a $1$-variation path with $\|t\|_{1\text{-var};[0,1]} = 1$. We apply the level-$\lfloor\gamma\rfloor$ Euler scheme to this ODE: $\mathcal{E}^{\lfloor\gamma\rfloor}_{0,1}(y_s; F^{\lfloor\gamma\rfloor}_{s,t}, t)$ is the level-$\lfloor\gamma\rfloor$ truncated-signature step for the autonomous vector field $F^{\lfloor\gamma\rfloor}_{s,t}$ on the elementary bounded-variation driver $t$. For bounded-variation drivers, the [Local Error of Euler Scheme](/theorems/2537) specialises to the classical Taylor-expansion error: there exists $C_1 = C_1(\gamma)$ such that \begin{align*} \|u(1) - \mathcal{E}^{\lfloor\gamma\rfloor}_{0,1}(y_s; F^{\lfloor\gamma\rfloor}_{s,t}, t)\| \leq C_1\,\|F^{\lfloor\gamma\rfloor}_{s,t}\|_{\mathrm{Lip}^\gamma}^\gamma\,\|t\|_{1\text{-var};[0,1]}^\gamma. \end{align*} We verify the hypotheses of the cited bound: (i) $F^{\lfloor\gamma\rfloor}_{s,t}$ is $\mathrm{Lip}^\gamma$ — established in Step 1 — and (ii) $t$ is bounded-variation with $1$-variation equal to $1$. Both are satisfied. Substituting $\|t\|_{1\text{-var};[0,1]} = 1$ collapses the right-hand side to $C_1\,\|F^{\lfloor\gamma\rfloor}_{s,t}\|_{\mathrm{Lip}^\gamma}^\gamma$. The crucial observation: $u(1) = \mathcal{L}^{\lfloor\gamma\rfloor}_{s,t}(y_s; f, x)$ by the definition of the log-ODE method. So the bound becomes \begin{align*} \|\mathcal{L}^{\lfloor\gamma\rfloor}_{s,t}(y_s; f, x) - \mathcal{E}^{\lfloor\gamma\rfloor}_{0,1}(y_s; F^{\lfloor\gamma\rfloor}_{s,t}, t)\| \leq C_1\,\|F^{\lfloor\gamma\rfloor}_{s,t}\|_{\mathrm{Lip}^\gamma}^\gamma. \end{align*} Substituting the Step 1 bound $\|F^{\lfloor\gamma\rfloor}_{s,t}\|_{\mathrm{Lip}^\gamma} \leq \|f\|_{\mathrm{Lip}^\gamma}\|x\|_{p\text{-var};[s,t]}$ (modulo a $p,\gamma$-dependent constant) and absorbing all such constants into a relabelled $C_1 = C_1(p,\gamma)$ gives the form recorded above. The strategy of comparing the log-ODE solution to its own level-$\lfloor\gamma\rfloor$ Euler step on the auxiliary ODE is the linchpin: it lets us re-route through Euler-scheme estimates and ultimately connect to the original RDE Euler step in Step 3.[/guided]

[step:Identify the Euler step on the auxiliary ODE with the Euler step on the original RDE]We claim \begin{align*} \mathcal{E}^{\lfloor\gamma\rfloor}_{0,1}(y_s;\, F^{\lfloor\gamma\rfloor}_{s,t},\, t) = \mathcal{E}^{\lfloor\gamma\rfloor}_{s,t}(y_s;\, f,\, x). \end{align*} By definition, the Euler scheme of order $\lfloor\gamma\rfloor$ for an RDE driven by a $p$-rough path $x$ between times $s, t$ is \begin{align*} \mathcal{E}^{\lfloor\gamma\rfloor}_{s,t}(y_s;\, f,\, x) = \Phi_f\!\big(S^{\lfloor\gamma\rfloor}(x)_{s,t}\big)(y_s), \end{align*} where $S^{\lfloor\gamma\rfloor}(x)_{s,t}$ is the level-$\lfloor\gamma\rfloor$ truncated signature of $x|_{[s,t]}$, and $\Phi_f$ extends from Lie polynomials to the truncated tensor algebra as a homomorphism. The truncated signature and truncated log-signature are related by the tensor-exponential identity \begin{align*} S^{\lfloor\gamma\rfloor}(x)_{s,t} = \exp^{\otimes\lfloor\gamma\rfloor}\!\big(\log^{\lfloor\gamma\rfloor}(S(x)_{s,t})\big), \end{align*} where $\exp^{\otimes\lfloor\gamma\rfloor}$ is the truncated tensor exponential, defined on $\mathfrak{L}^{\lfloor\gamma\rfloor}$ as the polynomial $\xi \mapsto \sum_{k=0}^{\lfloor\gamma\rfloor} \xi^{\otimes k}/k!$. Because $\Phi_f$ is a Lie-algebra homomorphism that respects the tensor-algebra structure up to truncation, it intertwines the tensor exponential with the time-1 flow of vector fields. Concretely, \begin{align*} \Phi_f\!\big(\exp^{\otimes\lfloor\gamma\rfloor}(\xi)\big)(y_s) = \mathcal{E}^{\lfloor\gamma\rfloor}_{0,1}(y_s;\, \Phi_f(\xi),\, t) \end{align*} for any $\xi \in \mathfrak{L}^{\lfloor\gamma\rfloor}(\mathbb{R}^d)$, where the right-hand side is the level-$\lfloor\gamma\rfloor$ Euler step on $[0,1]$ for the autonomous ODE driven by the linear path $r \mapsto r$. Setting $\xi = \log^{\lfloor\gamma\rfloor}(S(x)_{s,t})$ and recalling $F^{\lfloor\gamma\rfloor}_{s,t} = \Phi_f(\xi)$, \begin{align*} \mathcal{E}^{\lfloor\gamma\rfloor}_{s,t}(y_s;\, f,\, x) = \Phi_f(S^{\lfloor\gamma\rfloor}(x)_{s,t})(y_s) = \Phi_f(\exp^{\otimes\lfloor\gamma\rfloor}(\xi))(y_s) = \mathcal{E}^{\lfloor\gamma\rfloor}_{0,1}(y_s;\, F^{\lfloor\gamma\rfloor}_{s,t},\, t). \end{align*} This is the asserted equality.[/step]

[guided]We have the bound from Step 2: \begin{align*} \|\mathcal{L}^{\lfloor\gamma\rfloor}_{s,t}(y_s;\, f,\, x) - \mathcal{E}^{\lfloor\gamma\rfloor}_{0,1}(y_s;\, F^{\lfloor\gamma\rfloor}_{s,t},\, t)\| \leq C_1\big(\|f\|_{\mathrm{Lip}^\gamma}\|x\|_{p\text{-var};[s,t]}\big)^\gamma. \end{align*} The right-hand side has $\|y_t - \mathcal{L}^{\lfloor\gamma\rfloor}_{s,t}(y_s)\|$ as our target, and we want to bring in the level-$\lfloor\gamma\rfloor$ Euler scheme $\mathcal{E}^{\lfloor\gamma\rfloor}_{s,t}(y_s; f, x)$ on the original signal so we can use the existing *Local Error of Euler Scheme* result. The trick is to note that the auxiliary Euler step $\mathcal{E}^{\lfloor\gamma\rfloor}_{0,1}(y_s; F^{\lfloor\gamma\rfloor}_{s,t}, t)$ and the original Euler step $\mathcal{E}^{\lfloor\gamma\rfloor}_{s,t}(y_s; f, x)$ are **literally the same point** in $\mathbb{R}^e$, even though they look different. Why? Both Euler schemes can be written in the same algebraic form: \begin{align*} \mathcal{E}^{\lfloor\gamma\rfloor}_{s,t}(y_s; f, x) &= \Phi_f\!\big(S^{\lfloor\gamma\rfloor}(x)_{s,t}\big)(y_s), \\ \mathcal{E}^{\lfloor\gamma\rfloor}_{0,1}(y_s; F^{\lfloor\gamma\rfloor}_{s,t}, t) &= \Phi_{F^{\lfloor\gamma\rfloor}_{s,t}}\!\big(S^{\lfloor\gamma\rfloor}(t)_{0,1}\big)(y_s) = \Phi_f\!\big(\exp^{\otimes\lfloor\gamma\rfloor}(\xi)\big)(y_s), \end{align*} where $\xi = \log^{\lfloor\gamma\rfloor}(S(x)_{s,t})$. The second equality in the second line uses that $S^{\lfloor\gamma\rfloor}(t)_{0,1}$ along the linear path $r \mapsto r$ is just the elementary truncated signature whose action via $\Phi_{F^{\lfloor\gamma\rfloor}_{s,t}}$ corresponds to applying $F^{\lfloor\gamma\rfloor}_{s,t}$ as a single time-1 flow on the auxiliary ODE — and the homomorphism property of $\Phi_f$ then collapses this back to $\Phi_f(\exp^{\otimes\lfloor\gamma\rfloor}(\xi))$. The decisive identity is $S^{\lfloor\gamma\rfloor}(x)_{s,t} = \exp^{\otimes\lfloor\gamma\rfloor}(\log^{\lfloor\gamma\rfloor}(S(x)_{s,t}))$, the **truncated log-signature inversion**. Both the truncated tensor exponential and the truncated log are polynomial maps in the truncated tensor algebra, and they are mutual inverses on group-like elements (which is what truncated signatures of geometric rough paths are). Hence the two Euler steps coincide: \begin{align*} \mathcal{E}^{\lfloor\gamma\rfloor}_{s,t}(y_s;\, f,\, x) = \mathcal{E}^{\lfloor\gamma\rfloor}_{0,1}(y_s;\, F^{\lfloor\gamma\rfloor}_{s,t},\, t). \end{align*} This is the algebraic identity that makes the log-ODE method work: solving the log-ODE on $[0,1]$ with vector field $F^{\lfloor\gamma\rfloor}_{s,t}$ to high order is the **same** as taking a single Euler step on the original RDE.[/guided]

[step:Apply the local error of Euler scheme to bound $\|y_t - \mathcal{E}^{\lfloor\gamma\rfloor}_{s,t}(y_s; f, x)\|$]By the *Local Error of Euler Scheme* theorem, with hypotheses verified — $x \in G\Omega_p(\mathbb{R}^d)$ a geometric $p$-rough path, $f \in \mathrm{Lip}^\gamma$ with $\gamma > p$, and $y$ the RDE solution with initial condition $y_s$ — there exists $C_2 = C_2(p, \gamma)$ such that \begin{align*} \|y_t - \mathcal{E}^{\lfloor\gamma\rfloor}_{s,t}(y_s;\, f,\, x)\| \leq C_2\,\big(\|f\|_{\mathrm{Lip}^\gamma}\,\|x\|_{p\text{-var};[s,t]}\big)^\gamma. \end{align*}[/step]

[guided]We invoke the [Local Error of Euler Scheme](/theorems/2537) to bound the gap between the true RDE solution $y_t$ and its level-$\lfloor\gamma\rfloor$ Euler one-step approximation $\mathcal{E}^{\lfloor\gamma\rfloor}_{s,t}(y_s; f, x)$. The cited theorem gives: for any geometric $p$-rough path $x \in G\Omega_p(\mathbb{R}^d)$, $\mathrm{Lip}^\gamma$ vector field $f$ with $\gamma > p$, and RDE solution $y$ with $y_s$ at time $s$, there exists $C_2 = C_2(p, \gamma)$ such that \begin{align*} \|y_t - \mathcal{E}^{\lfloor\gamma\rfloor}_{s,t}(y_s;\, f,\, x)\| \leq C_2\,\big(\|f\|_{\mathrm{Lip}^\gamma}\,\|x\|_{p\text{-var};[s,t]}\big)^\gamma. \end{align*} We verify the three hypotheses of the cited theorem: - *Geometric rough path:* $x \in G\Omega_p(\mathbb{R}^d)$ is the standing hypothesis of the present theorem. - *Vector field regularity:* $f \in \mathrm{Lip}^\gamma$ with $\gamma > p$ is the standing hypothesis (the log-ODE method requires the same threshold as the RDE itself). - *RDE solution:* $y$ is the unique RDE solution started at $y_s$ at time $s$, which exists and is unique because the Universal Limit Theorem applies under the verified $\mathrm{Lip}^\gamma$ regularity with $\gamma > p$. The bound is *not* the same as the bound in Step 2 — it compares the true RDE to its Euler step (with the original signal $x$ on the original interval $[s,t]$), whereas Step 2 compares the log-ODE solution to its Euler step (on the auxiliary signal $t$ on $[0,1]$). The Step 3 algebraic identity tells us these two Euler steps are *the same point*, so when we triangle-inequality them together in Step 5, the Step 2 and Step 4 bounds can be added directly.[/guided]

[step:Conclude via the triangle inequality] Combining Step 3 (which lets us replace $\mathcal{E}^{\lfloor\gamma\rfloor}_{0,1}(y_s; F^{\lfloor\gamma\rfloor}_{s,t}, t)$ with $\mathcal{E}^{\lfloor\gamma\rfloor}_{s,t}(y_s; f, x)$ in the bound from Step 2) with Step 4 and the triangle inequality, \begin{align*} \|y_t - \mathcal{L}^{\lfloor\gamma\rfloor}_{s,t}(y_s;\, f,\, x)\| &\leq \|y_t - \mathcal{E}^{\lfloor\gamma\rfloor}_{s,t}(y_s;\, f,\, x)\| + \|\mathcal{E}^{\lfloor\gamma\rfloor}_{s,t}(y_s;\, f,\, x) - \mathcal{L}^{\lfloor\gamma\rfloor}_{s,t}(y_s;\, f,\, x)\| \\ &\leq C_2\,\big(\|f\|_{\mathrm{Lip}^\gamma}\,\|x\|_{p\text{-var};[s,t]}\big)^\gamma + C_1\,\big(\|f\|_{\mathrm{Lip}^\gamma}\,\|x\|_{p\text{-var};[s,t]}\big)^\gamma \\ &= (C_1 + C_2)\,\big(\|f\|_{\mathrm{Lip}^\gamma}\,\|x\|_{p\text{-var};[s,t]}\big)^\gamma. \end{align*} Setting $C := C_1 + C_2 = C(p, \gamma)$ gives the claimed estimate \begin{align*} \|y_t - \mathcal{L}^{\lfloor\gamma\rfloor}_{s,t}(y_s;\, f,\, x)\| \leq C\,\big(\|f\|_{\mathrm{Lip}^\gamma}\,\|x\|_{p\text{-var};[s,t]}\big)^\gamma, \end{align*} completing the proof. [/step]