Of course it is childish, but not incorrect. And furthermore, it made ou turn the right way. I have given up any representation of quantum mechanics, I am only handling the real quantum mechanical quantities. But you don't seem to like that either.
And for this purpose, you don't need the comutator rules for observables, for two main reasons:
a)First of all,the moment you say "order of measurments"(and you can find your own quote on this one some 30 messages ago) the commutation of the observables M1 and M2 go out the window, because you are dealing with operators at different instants in time, and wavefunctions, etc, while the commutation of operators are equal time commutation rules. You should already know this from QFT and Feynmann graphs.
b)Second of all, the practice of measurement, no one in QM "measures" simultaneously anything, unless some experiment is specifically designed that way. So from the praxis point of viev, the commutation of observable is irrelevant.
But then let's return to your argument, once again. You keep the observables as operators, they commute, but you don't want any projection postulate involeved.Your proof can be summarized as follows: <S|M1M2|S>=<S|M2M1|S> which principially can tell you nothing about entanglement, even in this context, since you are measuring simultaneously the two quantieties/observables, so no ordering comes into play. At best it becomes an argument against non-locality, but not against entanglement.Once you take ordering into consideration, commutation fails (you have observables at different moment in time, so you cannot say anything about their commutation), and your argument becomes invalid, once again.
