[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 $ab \in S^{m+m'}(U \times \mathbb{R}^n)$.
[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]
[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]
