[proofplan]
We prove that $r$ lies in every negative integer order symbol class. Given an arbitrary order $-M$, the convergence $m_N \to -\infty$ lets us choose an index for which $m_N \leq -M$. The monotonicity of the symbol order filtration then embeds $S^{m_N}_{1,0}$ into $S^{-M}_{1,0}$, and intersecting over all $M$ gives $S^{-\infty}$. The special case $r \sim 0$ follows because the zero asymptotic expansion gives membership in the classes $S^{m_0-N}_{1,0}$, whose orders tend to $-\infty$.
[/proofplan]
[step:Lower the symbolic order below an arbitrary negative integer]
Fix $M \in \mathbb{N}$. Since the sequence $(m_N)_{N \in \mathbb{N}}$ satisfies $m_N \to -\infty$, there exists $N_M \in \mathbb{N}$ such that $m_{N_M} \leq -M$. By hypothesis,
\begin{align*}
r \in S^{m_{N_M}}_{1,0}(U \times \mathbb{R}^n).
\end{align*}
The symbol order filtration is monotone in the order parameter: if $a \leq b$, then
\begin{align*}
S^a_{1,0}(U \times \mathbb{R}^n) \subset S^b_{1,0}(U \times \mathbb{R}^n).
\end{align*}
Applying this with $a := m_{N_M}$ and $b := -M$ gives
\begin{align*}
S^{m_{N_M}}_{1,0}(U \times \mathbb{R}^n) \subset S^{-M}_{1,0}(U \times \mathbb{R}^n).
\end{align*}
Therefore
\begin{align*}
r \in S^{-M}_{1,0}(U \times \mathbb{R}^n).
\end{align*}
[guided]
The goal is to show that $r$ has arbitrarily negative symbolic order. So we begin with an arbitrary positive integer $M \in \mathbb{N}$ and aim to prove
\begin{align*}
r \in S^{-M}_{1,0}(U \times \mathbb{R}^n).
\end{align*}
The hypothesis gives a sequence of real orders $(m_N)_{N \in \mathbb{N}}$ with $m_N \to -\infty$. This means that the orders eventually fall below every prescribed negative bound. In particular, for the chosen $M$, there exists an index $N_M \in \mathbb{N}$ such that
\begin{align*}
m_{N_M} \leq -M.
\end{align*}
By the assumed membership condition on the same sequence of orders, we have
\begin{align*}
r \in S^{m_{N_M}}_{1,0}(U \times \mathbb{R}^n).
\end{align*}
Now we use the monotonicity built into the symbol classes. A symbol estimate of lower order is stronger than a symbol estimate of higher order: if $a \leq b$, then every symbol in $S^a_{1,0}(U \times \mathbb{R}^n)$ also belongs to $S^b_{1,0}(U \times \mathbb{R}^n)$. Applying this with $a := m_{N_M}$ and $b := -M$, the inequality $m_{N_M} \leq -M$ gives
\begin{align*}
S^{m_{N_M}}_{1,0}(U \times \mathbb{R}^n) \subset S^{-M}_{1,0}(U \times \mathbb{R}^n).
\end{align*}
Combining the membership of $r$ in the smaller class with this inclusion yields
\begin{align*}
r \in S^{-M}_{1,0}(U \times \mathbb{R}^n).
\end{align*}
[/guided]
[/step]
[step:Intersect over all negative integer orders]
The integer $M \in \mathbb{N}$ was arbitrary. Hence
\begin{align*}
r \in \bigcap_{M=1}^{\infty} S^{-M}_{1,0}(U \times \mathbb{R}^n).
\end{align*}
By the definition of the smoothing symbol class,
\begin{align*}
S^{-\infty}(U \times \mathbb{R}^n) := \bigcap_{M=1}^{\infty} S^{-M}_{1,0}(U \times \mathbb{R}^n).
\end{align*}
Therefore
\begin{align*}
r \in S^{-\infty}(U \times \mathbb{R}^n).
\end{align*}
[/step]
[step:Apply the criterion to a symbol with zero asymptotic expansion]
Assume there exists $m_0 \in \mathbb{R}$ such that $r \in S^{m_0}_{1,0}(U \times \mathbb{R}^n)$ and $r \sim 0$ in the stated sense. By the definition of zero asymptotic expansion used here, for every $N \in \mathbb{N}$ one has
\begin{align*}
r \in S^{m_0-N}_{1,0}(U \times \mathbb{R}^n).
\end{align*}
Define the sequence $(\ell_N)_{N \in \mathbb{N}}$ by $\ell_N := m_0-N$. Then $\ell_N \to -\infty$ as $N \to \infty$, and the preceding criterion applies with $m_N := \ell_N$. Hence
\begin{align*}
r \in S^{-\infty}(U \times \mathbb{R}^n).
\end{align*}
Thus $r$ is smoothing.
[/step]
