If $[M,N]=0$, is the probability distribution obtained measuring $M$ after $N$ the same as the one obtained measuring $N$?

Let $$|\psi\rangle$$ be a fixed state and $$M$$ and $$N$$ two commuting operators corresponding to projective measurements. Consider the probability distribution $$p$$ obtained by measuring $$N$$ on $$|\psi\rangle$$. Is this probability distribution exactly the same as the probability distribution $$q$$ that we would get if we measured first $$M$$ on $$|\psi\rangle$$, discarded the result, then measured $$N$$ on the post-measurement state?

I managed to prove it using the basic law of total probability and the existence of a common orthogonal basis, but I want to be sure since I have never seen it.

Let's give a decomposition of both operators in terms of projectors onto their eigenspaces $$N=\sum_{q}n_{q}Q_{q},\qquad M=\sum_{p}m_{p}P_{p}.$$ The probability of getting measurement result $$q$$ on measuring $$N$$ is $$p_q=\text{Tr}(Q_q|\psi\rangle\langle\psi|).$$
Instead, imagine we first measure $$M$$ and discard the result. After that measurement, we have $$\rho=\sum_pP_p|\psi\rangle\langle\psi|P_p.$$ If we now measure $$N$$, the probability of getting result $$q$$ is $$\tilde{p}_q=\text{Tr}(Q_q\rho).$$ Using the fact that $$[Q_q,P_p]=0$$, this can be written as $$\tilde{p}_q=\text{Tr}(\sum_pP_pQ_q|\psi\rangle\langle\psi|P_p).$$ Now use the cyclic property of trace, and the fact that $$P_p^2=P_p$$ to give $$\tilde p_q=\text{Tr}\left(\left(\sum_pP_p\right)Q_q|\psi\rangle\langle\psi|\right).$$ Since $$\sum_pP_p=I$$, this shows that $$p_q=\tilde p_q$$, as required.