Operator splitting vs Predictor corrector

For simplicity, lets assume $F_1$ is solved by doing the Hyman method and $F_2$ the first order time derivative. Now, lets advance in time using operator splitting. In other words, we are doing the following:

$$u^{n+(1/3+1/2)/2} = u^{n} + a_1(u^{n-1}-u^{n}) + b_1 F_1 u^{n}$$

$$u^{n+(2/3+1/2)/2} = u^{n+(1/3+1/2)/2} + a_2(u^n-u^{n+(1/3+1/2)/2}) + b_2 F_1 u^{n}$$

$$u^{n+(1+1/2)/2} = u^{n+(2/3+1/2)/2} + c F_1 u^{n+(2/3+1/2)/2}$$

$$u^{n+1} = u^{n+(1+1/2)/2} + dt F_2 u^{n+(1+1/2)/2}$$

Note that the last step is the 1st order timestep for the second operator ($F_2$). The super-index means (full-time-steps + fraction of step from Hyman + fraction of step from the operator splitting). Now, lets simplify this problem and assume that $F_1,u = 0$. One can collapse all the steps XX-YY as follows:

$$u^{n+1} =u^{n} +a_2(a_1(u^{n-1}-u^{n})) +a_1(u^{n-1}-u^{n}) + dt F_2 u^{n+(1+1/2)/2}$$

Obviously, this solution is wrong by $a_2(a_1(u^{n-1}-u^{n}))+a_1(u^{n-1}-u^{n})$. In a very simplified scenario, where $F_2$ is constant in time, then the second term is adding in each step $a_2(a_1(u^{n-1}-u^{n}))+a_1(u^{n-1}-u^{n})$, which is going to be $-a_2(a_1,F_2,u^{n-1})+a_1F_2u^{n-1}$ coming from the previous time step.