Let $X$ be a cosimplicial space and suppose that:
the matching maps $$ s:X^{k}\rightarrow M^{k-1}X $$ are fibrations for all $k\leq n$.
On one hand we can form the partial totalization $Tot_n(X)$, on the other hand we can take the Reedy fibrant replacement $RX$ and define the homotopy invariant totalization $\widetilde{Tot}_n(X):=Tot_n(RX)$.
Under condition above, do we have $Tot_n(X)\simeq \widetilde{Tot}_n(X)$?
In other words, is the partial totalization in this case already homotopy invariant?
For example, totalization preserves weak equivalences between Reedy fibrant cosimplicial spaces (i.e. those for which all the matching maps are fibrations): if $X$ and $Y$ are (levelwise) weakly equivalent and both are Ready fibrant, then $Tot(X)\simeq Tot(Y)$.
I guess I am asking if there is an analogous statement for the case of partial totalizations, and would it be as simple as I suggested - partial totalization preserves weak equivalences between truncated cosimplicial spaces if they are ''fibrant up to appropriate degree''.
EDIT:
I realized that one has: $$ Tot_n(X)\simeq Tot(cosk_nX) $$ where $cosk_nX$ is the coskeleton of the cosimplicial space $X$, i.e. the composition of the n-th truncation functor with its right Kan extension. See e.g. Eldred's notes, at the bottom of this page.
So I think my question boils down to:
If $X$ satisfies the condition above, is $cosk_nX$ Reedy fibrant?