[proofplan]
The proof is local in coordinate charts, because the principal symbol and the principal part of the commutator are computed in the local semiclassical symbol calculus. In a local quantization, the composition formula shows that the symbols of $AB$ and $BA$ have the same product term $ab$, so the order $m+m'$ contribution cancels. The antisymmetric part of the first correction term is exactly
\begin{align*}
\frac{h}{i}\{a,b\},
\end{align*}
and all remaining terms are of lower symbolic order after one factor of $h$ is extracted. Finally, the local expressions glue because the Poisson bracket is defined intrinsically by the canonical symplectic structure on $T^*M$.
[/proofplan]
[step:Reduce the commutator calculation to a local symbol computation]
Let $(U,\varphi)$ be a coordinate chart on $M$, with induced canonical coordinates $(x,\xi)$ on $T^*U$. Let
\begin{align*}
\operatorname{Op}_h: S^r(T^*U) \to \Psi_h^r(U)
\end{align*}
denote the chosen local semiclassical quantization map in this chart, with the convention that the first-order term in the local composition formula for symbols $p_h \in S^r(T^*U)$ and $q_h \in S^s(T^*U)$ is
\begin{align*}
\frac{h}{i}\sum_{j=1}^n \partial_{\xi_j}p\,\partial_{x_j}q,
\end{align*}
where $p$ and $q$ are the principal parts of $p_h$ and $q_h$, respectively. Choose local full symbols
\begin{align*}
a_h \in S^m(T^*U)
\end{align*}
and
\begin{align*}
b_h \in S^{m'}(T^*U)
\end{align*}
for $A$ and $B$, respectively, so that
\begin{align*}
a_h = a + h a_1
\end{align*}
with $a_1 \in S^{m-1}(T^*U)$, and
\begin{align*}
b_h = b + h b_1
\end{align*}
with $b_1 \in S^{m'-1}(T^*U)$, modulo symbols of lower semiclassical order.
Since $A$ and $B$ are properly supported, the compositions $AB$ and $BA$ are well-defined in the semiclassical calculus. It is enough to compute the principal local symbol of $AB-BA$ on each coordinate chart and then check that the resulting expression is invariant under changes of coordinates.
[/step]
[step:Use the composition formula to isolate the antisymmetric first correction]
By the local semiclassical composition formula, the full symbol of $AB$ is, modulo $h^2 S^{m+m'-2}(T^*U)$,
\begin{align*}
a_h \# b_h = a b + \frac{h}{i} \sum_{j=1}^n \partial_{\xi_j} a \, \partial_{x_j} b + h(a_1 b + a b_1).
\end{align*}
Similarly, the full symbol of $BA$ is, modulo $h^2 S^{m+m'-2}(T^*U)$,
\begin{align*}
b_h \# a_h = b a + \frac{h}{i} \sum_{j=1}^n \partial_{\xi_j} b \, \partial_{x_j} a + h(b_1 a + b a_1).
\end{align*}
The zeroth-order products cancel because $ab=ba$. The terms involving $a_1$ and $b_1$ also cancel because multiplication of scalar symbols is commutative:
\begin{align*}
a_1 b + a b_1 - b_1 a - b a_1 = 0.
\end{align*}
Therefore the full local symbol of $[A,B]$ is, modulo $h^2 S^{m+m'-2}(T^*U)$,
\begin{align*}
\frac{h}{i} \sum_{j=1}^n \left(\partial_{\xi_j} a \, \partial_{x_j} b - \partial_{\xi_j} b \, \partial_{x_j} a\right).
\end{align*}
[guided]
The composition formula is where the factor of $h$ and the Poisson bracket enter. In a local semiclassical quantization, composing two operators corresponds to multiplying their symbols with a noncommutative product, denoted $\#$. Up to the first power of $h$, this product has the form
\begin{align*}
a_h \# b_h = a b + \frac{h}{i} \sum_{j=1}^n \partial_{\xi_j} a \, \partial_{x_j} b + h(a_1 b + a b_1)
\end{align*}
modulo $h^2 S^{m+m'-2}(T^*U)$, where $a_1$ and $b_1$ are the first lower-order symbol coefficients of $A$ and $B$.
Now reverse the order. The symbol of $BA$ is
\begin{align*}
b_h \# a_h = b a + \frac{h}{i} \sum_{j=1}^n \partial_{\xi_j} b \, \partial_{x_j} a + h(b_1 a + b a_1)
\end{align*}
modulo the same lower class $h^2 S^{m+m'-2}(T^*U)$. The commutator subtracts these two expressions. The product terms satisfy $ab-ba=0$, so there is no order $m+m'$ contribution. This is the reason the commutator gains one semiclassical factor and one symbolic order.
The lower coefficient terms also cancel:
\begin{align*}
a_1 b + a b_1 - b_1 a - b a_1 = 0.
\end{align*}
Thus the only first-order contribution that survives is the antisymmetric derivative term:
\begin{align*}
\frac{h}{i} \sum_{j=1}^n \left(\partial_{\xi_j} a \, \partial_{x_j} b - \partial_{\xi_j} b \, \partial_{x_j} a\right).
\end{align*}
This is precisely the local coordinate formula for the Poisson bracket, written with the sign convention
\begin{align*}
\{a,b\} = \sum_{j=1}^n \left(\partial_{\xi_j} a \, \partial_{x_j} b - \partial_{\xi_j} b \, \partial_{x_j} a\right).
\end{align*}
Therefore the local principal contribution to the commutator is $\frac{h}{i}\{a,b\}$.
[/guided]
[/step]
[step:Identify the symbolic order of the remainder]
Let $c_h \in S^{m+m'}(T^*U)$ denote a local full symbol of the commutator $[A,B]$ in the chosen chart, so that $[A,B]$ is locally represented by $\operatorname{Op}_h(c_h)$ modulo residual operators. The displayed computation gives
\begin{align*}
c_h - \frac{h}{i}\{a,b\} \in h^2 S^{m+m'-2}(T^*U).
\end{align*}
Since $\{a,b\} \in S^{m+m'-1}(T^*U)$, the leading term $\frac{h}{i}\{a,b\}$ lies in $hS^{m+m'-1}(T^*U)$. The remainder lies in $h^2S^{m+m'-2}(T^*U)$, which is precisely the next lower class for the principal symbol map on $h\Psi_h^{m+m'-1}(U)$. Hence, locally,
\begin{align*}
[A,B] \in h\Psi_h^{m+m'-1}(U)
\end{align*}
and its principal symbol as an element of that class is represented by $\frac{h}{i}\{a,b\}$.
[/step]
[step:Glue the local principal symbols using the canonical Poisson bracket]
The local expression
\begin{align*}
\{a,b\} = \sum_{j=1}^n \left(\partial_{\xi_j} a \, \partial_{x_j} b - \partial_{\xi_j} b \, \partial_{x_j} a\right)
\end{align*}
is the coordinate formula for the Poisson bracket determined by the canonical symplectic form on $T^*M$. Because this symplectic form is intrinsic, the leading local expressions computed in overlapping coordinate charts agree on intersections as representatives of the principal symbol. The coordinate-dependent lower-order full-symbol representatives may differ, but their differences lie in the quotient class absorbed by $h^2S^{m+m'-2}$. Therefore the local principal symbols glue to the global symbol
\begin{align*}
\frac{h}{i}\{a,b\} \in hS^{m+m'-1}(T^*M) / h^2 S^{m+m'-2}(T^*M).
\end{align*}
Residual terms remain residual under subtraction, and lower symbolic terms are absorbed into $h^2\Psi_h^{m+m'-2}(M)$. Consequently
\begin{align*}
[A,B] \in h\Psi_h^{m+m'-1}(M)
\end{align*}
with principal symbol
\begin{align*}
\sigma_h^{m+m'-1}([A,B]) = \frac{h}{i}\{a,b\}.
\end{align*}
This proves the theorem.
[/step]
