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.