Consider the following initial value problem $$\begin{cases} x'(t) = f(t,x(t)), \; t \in [t_0, T]\\ x(t_0) = \mu_0 \end{cases}$$ where $f \in \mathcal{C}\left([t_0, T] \times \mathbb{R}^n;\mathbb{R}^n \right)$ and is lipschitz on the second variable with Lipschitz's constant $L >0$. Given $n \in \mathbb{N}$, consider the net of nodes $t_i = t_0 + ih$ for $i = 0,1,\dots , N$ where $h = \frac{T-t_0}{N}$
Now consider the predictor-corrects method $P(EC)E$ where the predictor is the modified Euler Method $$x_{i+1} = x_i + hf\left( t_i + \frac{h}{2}, x_i + \frac{h}{2}f(t_i,x_i)\right)$$ and the corrector is the two step Adams- Moulton Method given by $$x_{i+2} = x_{i+1} + \frac{h}{12}\left( 5f_{i+2} + 8f_{i+1} - f_i \right)$$ where $f_{i+k} = f(t_{i+k}, x_{i+k})$
Show that the order of the resultant predictor-corrector method is $3$.
Now, I've already shown that the Adam-Mouton's Method has order $3$, and that Euler's modifies method has order two. Now there's a Theorem which barely says that
If I have two methods, both linear multistep methods of orders $p$ and $p^*$, the resultant predictor-corrector method taking as the predictor the method of order $p$ and the corrector the method of order $p^+$ has order $$\overline{p}=\min \{p,p^*\}$$
However, Euler's modified method is a Runge-Kutta method and therefore I don't think this theorem applies. My question is
How do I find the order of the method?
Thank you in advance for your help.
Compare to the exact solution $x(t)$ of the ODE going through the initial point $(t_i,x_i)$.
The improved Euler step then relates to the exact ODE solution as $$x_{i+1}=x(t_{i+1})+O(h^3)$$ and thus also $$f_{i+1}=f(t_{i+1},x_{i+1})=x'(t_{i+1})+O(h^3).$$
The error of the Adams-Moulton step is $$ x(t_{i+2})=x(t_{i+1})+\frac{h}{12}(5x'(t_{i+2})+8x'(t_{i+1})-x'(t_i))+O(h^4) $$ Replacing $x'(t_{i+1})$ with $f_{i+1}$ only adds another $h·O(h^3)=O(h^4)$ error term. As we know that $x(t_{i+2})$ nearly solves the fixed-point equation for $x_{i+2}=X$, $$ X=C+\frac{5h}{12}f(t_{i+2},X),\qquad C=x_{i+1}+\frac{h}{12}(8f_{i+1}-f_i) $$ with an error $O(h^4)$, by the implicit function theorem or more basically the Banach fixed point theorem, the exact solution of the Adams-Moulton step also only has distance of $O(h^4)$ to $x(t_{i+2})$.
This local $O(h^4)$ error translates into a global $O(h^3)$ error, thus giving the order of the method as $3$.