[proofplan] The proof is a direct bookkeeping argument from the defining estimates of the standard symbol class $S^m_{1,0}$. Differentiating in $x$ only shifts the local seminorm index, while differentiating in $\xi$ lowers the symbol order by the number of $\xi$-derivatives. Products are handled by the multi-index Leibniz formula: each summand has one $\xi$-derivative split between the two factors, and the losses add to exactly $|\beta|$. Multiplication by $\psi(x)$ or $\chi(\xi)$ is the same product argument with an order-zero factor. [/proofplan]

[step:Shift the symbol estimates to prove stability under derivatives] Fix multi-indices $\alpha,\beta \in \mathbb{N}_0^n$, and define \begin{align*} c: U \times \mathbb{R}^n \to \mathbb{C} \end{align*} by \begin{align*} c(x,\xi) = \partial_x^\alpha \partial_\xi^\beta a(x,\xi). \end{align*} Since $a$ is smooth on $U \times \mathbb{R}^n$, the function $c$ is smooth on $U \times \mathbb{R}^n$. Let $K \subset U$ be compact, and let $\gamma,\delta \in \mathbb{N}_0^n$ be arbitrary multi-indices. Since $a \in S^m(U \times \mathbb{R}^n)$, there is a constant $C_{K,\alpha+\gamma,\beta+\delta} > 0$ such that \begin{align*} |\partial_x^{\alpha+\gamma}\partial_\xi^{\beta+\delta}a(x,\xi)| \leq C_{K,\alpha+\gamma,\beta+\delta}\langle \xi \rangle^{m-|\beta+\delta|} \end{align*} for all $(x,\xi) \in K \times \mathbb{R}^n$. Because $|\beta+\delta| = |\beta| + |\delta|$, this becomes \begin{align*} |\partial_x^\gamma\partial_\xi^\delta c(x,\xi)| \leq C_{K,\alpha+\gamma,\beta+\delta}\langle \xi \rangle^{m-|\beta|-|\delta|}. \end{align*} This is precisely the defining estimate for $c \in S^{m-|\beta|}(U \times \mathbb{R}^n)$. Hence \begin{align*} \partial_x^\alpha \partial_\xi^\beta a \in S^{m-|\beta|}(U \times \mathbb{R}^n). \end{align*} [/step]

[step:Apply the multi-index Leibniz formula to the product $ab$]Define \begin{align*} p: U \times \mathbb{R}^n \to \mathbb{C} \end{align*} by \begin{align*} p(x,\xi) = a(x,\xi)b(x,\xi). \end{align*} Since $a$ and $b$ are smooth on $U \times \mathbb{R}^n$, the product $p$ is smooth on $U \times \mathbb{R}^n$. Fix a compact set $K \subset U$ and multi-indices $\alpha,\beta \in \mathbb{N}_0^n$. The multi-index Leibniz formula gives \begin{align*} \partial_x^\alpha\partial_\xi^\beta p = \sum_{\alpha_1+\alpha_2=\alpha} \sum_{\beta_1+\beta_2=\beta} \binom{\alpha}{\alpha_1}\binom{\beta}{\beta_1} (\partial_x^{\alpha_1}\partial_\xi^{\beta_1}a) (\partial_x^{\alpha_2}\partial_\xi^{\beta_2}b). \end{align*} For each summand, the symbol estimates for $a$ and $b$ give constants $A_{K,\alpha_1,\beta_1} > 0$ and $B_{K,\alpha_2,\beta_2} > 0$ such that \begin{align*} |\partial_x^{\alpha_1}\partial_\xi^{\beta_1}a(x,\xi)| \leq A_{K,\alpha_1,\beta_1}\langle \xi \rangle^{m-|\beta_1|} \end{align*} and \begin{align*} |\partial_x^{\alpha_2}\partial_\xi^{\beta_2}b(x,\xi)| \leq B_{K,\alpha_2,\beta_2}\langle \xi \rangle^{m'-|\beta_2|}. \end{align*} Multiplying these estimates and using $|\beta_1|+|\beta_2|=|\beta|$ yields \begin{align*} |(\partial_x^{\alpha_1}\partial_\xi^{\beta_1}a)(x,\xi)(\partial_x^{\alpha_2}\partial_\xi^{\beta_2}b)(x,\xi)| \leq A_{K,\alpha_1,\beta_1}B_{K,\alpha_2,\beta_2} \langle \xi \rangle^{m+m'-|\beta|}. \end{align*} Since the Leibniz sum is finite, define \begin{align*} C_{K,\alpha,\beta} = \sum_{\alpha_1+\alpha_2=\alpha} \sum_{\beta_1+\beta_2=\beta} \binom{\alpha}{\alpha_1}\binom{\beta}{\beta_1} A_{K,\alpha_1,\beta_1}B_{K,\alpha_2,\beta_2}. \end{align*} Then \begin{align*} |\partial_x^\alpha\partial_\xi^\beta p(x,\xi)| \leq C_{K,\alpha,\beta}\langle \xi \rangle^{m+m'-|\beta|} \end{align*} for all $(x,\xi) \in K \times \mathbb{R}^n$. Therefore $ab \in S^{m+m'}(U \times \mathbb{R}^n)$.[/step]

[guided]We need to show that the product satisfies the same kind of estimates as a symbol of order $m+m'$. Define \begin{align*} p: U \times \mathbb{R}^n \to \mathbb{C} \end{align*} by \begin{align*} p(x,\xi) = a(x,\xi)b(x,\xi). \end{align*} Since $a$ and $b$ are smooth on $U \times \mathbb{R}^n$, the product $p$ is smooth on $U \times \mathbb{R}^n$. To prove $p \in S^{m+m'}(U \times \mathbb{R}^n)$, we fix a compact set $K \subset U$ and arbitrary multi-indices $\alpha,\beta \in \mathbb{N}_0^n$, and we must prove an estimate of the form \begin{align*} |\partial_x^\alpha\partial_\xi^\beta p(x,\xi)| \leq C_{K,\alpha,\beta}\langle \xi \rangle^{m+m'-|\beta|}. \end{align*} The only point to track is how the $\xi$-derivatives are distributed. The multi-index Leibniz formula gives \begin{align*} \partial_x^\alpha\partial_\xi^\beta p = \sum_{\alpha_1+\alpha_2=\alpha} \sum_{\beta_1+\beta_2=\beta} \binom{\alpha}{\alpha_1}\binom{\beta}{\beta_1} (\partial_x^{\alpha_1}\partial_\xi^{\beta_1}a) (\partial_x^{\alpha_2}\partial_\xi^{\beta_2}b). \end{align*} This formula is useful because the defining estimates for $S^m_{1,0}$ are stable under arbitrary $x$-derivatives but lose one order for each $\xi$-derivative. For a fixed summand, the hypothesis $a \in S^m(U \times \mathbb{R}^n)$ gives a constant $A_{K,\alpha_1,\beta_1} > 0$ such that \begin{align*} |\partial_x^{\alpha_1}\partial_\xi^{\beta_1}a(x,\xi)| \leq A_{K,\alpha_1,\beta_1}\langle \xi \rangle^{m-|\beta_1|}. \end{align*} Similarly, the hypothesis $b \in S^{m'}(U \times \mathbb{R}^n)$ gives a constant $B_{K,\alpha_2,\beta_2} > 0$ such that \begin{align*} |\partial_x^{\alpha_2}\partial_\xi^{\beta_2}b(x,\xi)| \leq B_{K,\alpha_2,\beta_2}\langle \xi \rangle^{m'-|\beta_2|}. \end{align*} Multiplying the two estimates gives \begin{align*} |(\partial_x^{\alpha_1}\partial_\xi^{\beta_1}a)(x,\xi)(\partial_x^{\alpha_2}\partial_\xi^{\beta_2}b)(x,\xi)| \leq A_{K,\alpha_1,\beta_1}B_{K,\alpha_2,\beta_2} \langle \xi \rangle^{m+m'-|\beta_1|-|\beta_2|}. \end{align*} Because $\beta_1+\beta_2=\beta$, we have $|\beta_1|+|\beta_2|=|\beta|$. Hence every summand has exactly the order \begin{align*} m+m'-|\beta|. \end{align*} There are only finitely many decompositions $\alpha_1+\alpha_2=\alpha$ and $\beta_1+\beta_2=\beta$, so we may define \begin{align*} C_{K,\alpha,\beta} = \sum_{\alpha_1+\alpha_2=\alpha} \sum_{\beta_1+\beta_2=\beta} \binom{\alpha}{\alpha_1}\binom{\beta}{\beta_1} A_{K,\alpha_1,\beta_1}B_{K,\alpha_2,\beta_2}. \end{align*} Adding the finitely many bounded summands then gives \begin{align*} |\partial_x^\alpha\partial_\xi^\beta p(x,\xi)| \leq C_{K,\alpha,\beta}\langle \xi \rangle^{m+m'-|\beta|} \end{align*} for all $(x,\xi) \in K \times \mathbb{R}^n$. This is the defining estimate for $p \in S^{m+m'}(U \times \mathbb{R}^n)$.[/guided]

[step:Treat multiplication by an $x$-factor as multiplication by an order-zero local symbol] Let $\psi \in C^\infty(U)$, and define \begin{align*} q: U \times \mathbb{R}^n \to \mathbb{C} \end{align*} by \begin{align*} q(x,\xi) = \psi(x)a(x,\xi). \end{align*} Since $\psi$ is smooth on $U$ and $a$ is smooth on $U \times \mathbb{R}^n$, the function $q$ is smooth on $U \times \mathbb{R}^n$. Fix a compact set $K \subset U$ and multi-indices $\alpha,\beta \in \mathbb{N}_0^n$. Since $\psi$ is smooth, for each $\alpha_1 \leq \alpha$ there is a finite constant \begin{align*} M_{K,\alpha_1} = \sup_{x \in K} |\partial_x^{\alpha_1}\psi(x)|. \end{align*} The Leibniz formula in the $x$-variables gives \begin{align*} \partial_x^\alpha\partial_\xi^\beta q = \sum_{\alpha_1+\alpha_2=\alpha} \binom{\alpha}{\alpha_1} (\partial_x^{\alpha_1}\psi) (\partial_x^{\alpha_2}\partial_\xi^\beta a). \end{align*} For each summand, the symbol estimate for $a$ gives \begin{align*} |(\partial_x^{\alpha_1}\psi)(x)(\partial_x^{\alpha_2}\partial_\xi^\beta a)(x,\xi)| \leq M_{K,\alpha_1}A_{K,\alpha_2,\beta}\langle \xi \rangle^{m-|\beta|}. \end{align*} Summing the finitely many terms gives the defining estimate for $q \in S^m(U \times \mathbb{R}^n)$. [/step]

[step:Treat multiplication by a $\xi$-cutoff as multiplication by an order-zero symbol] Let $\chi \in C^\infty(\mathbb{R}^n)$ satisfy the order-zero symbol estimates, and define \begin{align*} r: U \times \mathbb{R}^n \to \mathbb{C} \end{align*} by \begin{align*} r(x,\xi) = \chi(\xi)a(x,\xi). \end{align*} Since $\chi$ is smooth on $\mathbb{R}^n$ and $a$ is smooth on $U \times \mathbb{R}^n$, the function $r$ is smooth on $U \times \mathbb{R}^n$. Fix a compact set $K \subset U$ and multi-indices $\alpha,\beta \in \mathbb{N}_0^n$. Since $\chi$ is independent of $x$, the multi-index Leibniz formula in the $\xi$-variables gives \begin{align*} \partial_x^\alpha\partial_\xi^\beta r = \sum_{\beta_1+\beta_2=\beta} \binom{\beta}{\beta_1} (\partial_\xi^{\beta_1}\chi) (\partial_x^\alpha\partial_\xi^{\beta_2}a). \end{align*} For each summand, let $D_{\beta_1} > 0$ denote the constant in the order-zero estimate for $\chi$ with multi-index $\beta_1$. The order-zero estimate for $\chi$ and the order-$m$ estimate for $a$ give \begin{align*} |(\partial_\xi^{\beta_1}\chi)(\xi)(\partial_x^\alpha\partial_\xi^{\beta_2}a)(x,\xi)| \leq D_{\beta_1}A_{K,\alpha,\beta_2} \langle \xi \rangle^{-|\beta_1|} \langle \xi \rangle^{m-|\beta_2|}. \end{align*} Since $|\beta_1|+|\beta_2|=|\beta|$, this becomes \begin{align*} |(\partial_\xi^{\beta_1}\chi)(\xi)(\partial_x^\alpha\partial_\xi^{\beta_2}a)(x,\xi)| \leq D_{\beta_1}A_{K,\alpha,\beta_2} \langle \xi \rangle^{m-|\beta|}. \end{align*} The sum is finite, so the defining estimate for $r \in S^m(U \times \mathbb{R}^n)$ follows. This proves all stated stability properties of standard symbol classes. [/step]