pullback square in Goerss-Jardine In proving the the existence of the Reedy model structure on the category of simplicial objects in a model category $\mathcal{C}$, Goerss-Jardine prove there is a pullback square induced by a map of simplicial objects $f:X\to Y$, namely 
$$\begin{array}{ccc}
Y_n\times_{M_{n,k+1}}M_{n,k+1}X& \to & X_{n-1}\\
\downarrow & & \downarrow\\
Y_n\times_{M_{n,k}Y}M_{n,k}X& \to & Y_n\times_{M_{n-1,k}Y}M_{n-1,k}X
\end{array}$$
and I am having some trouble seeing exactly how they derive this pullback. For reference, this is lemma VII.2.2 in Goerss-Jardine. 
Here is what I have attempted so far. Here, Goerss-Jardine define
$$
M_{n,k}X = M_{\Delta^{n,k}}X
$$
where $\Delta^{n,k}$ is the simplicial set $d^0\Delta^{n-1}\cup \cdots \cup d^k\Delta^{n-1}$ and for a simplicial set $K$ and a simplicial object $X$ in $\mathcal{C}$, you can take
$$
M_KX:= \hom^{\Delta^\mathrm{op}}(K,X)
$$
From a standard pushout
$$\begin{array}{ccc}
\Delta^{n-1,k}& \to & \Delta^{n-1}\\
\downarrow & & \downarrow\\
\Delta^{n,k}& \to & \Delta^{n,k+1}
\end{array}$$
Since $M_{(-)}X$ takes pushouts to pullbacks, we get a pullback 
$$\begin{array}{ccc}
M_{n,k+1}X& \to & X_{n-1}\\
\downarrow & & \downarrow\\
M_{n,k}X& \to & M_{n-1,k}X
\end{array}$$
and we get a similar pullback for the partial matching objects on $Y$. My attempt at getting the desired pullback was to enlarge the diagram to the following 3-by-3 diagram
$$
\begin{array}{ccccc}
Y_n & \to & Y_{n-1} & \leftarrow & X_{n-1} \\
\downarrow & & \downarrow & & \downarrow \\
M_{n,k}Y & \to & M_{n-1,k}Y & \leftarrow & X_{n-1}\\
\uparrow & & \uparrow & & \uparrow \\
M_{n,k}X & \to & M_{n-1, k}X & \leftarrow & X_{n-1}
\end{array}
$$
The desired pullback diagram is then supposed to be obtained from this 3-by-3 diagram by taking the horizontal limits and then taking the corresponding pullback. I tried commuting these two limits, so I thought of taking the limits of the row and then taking the corresponding pullback, but this didn't really seem to simplify the problem. 
I am not sure if my attempt can be made to work and I am just not seeing something, or perhaps I am unaware of some category theory trick that Goerss-Jardine are using. Anyway, any hints would be greatly appreciated. 
 A: I would understand this proof as describing the limit of a "deleted 3-cube" in two different ways.  
We have a square involving maps $X_n\to X_{n-1}$, $X_n\to M_{n,k}X$, $X_{n-1}\to M_{n-1,k}X$, and $M_{n,k}X\to M_{n-1,k}X$; this square maps to a similar square with $X$ replaced with $Y$.  Altogether it is a cube.  I can't draw this cube here.
Now let's completely forget about the "maximal vertex" $X_n$.  The pullback you want to compute is actually the inverse limit of the deleted cube.  (It is also equal to the limit of your 3-by-3 diagram.)
When you compute the limit of a deleted 3-cube, you can slice it up in several different ways, so that it is equal to a pullback of three objects, one of which is a vertex, and the other two are pullbacks.  Goerss-Jardine have set it up one way.  But we can also slice it up as the limit of 
$$ Y_n \to Y_{n-1}\times_{M_{n-1,k}Y}M_{n,k}Y \leftarrow X_{n-1}\times_{M_{n-1,k}X} M_{n,k}X.
$$
GJ identify the rightmost term as $M_{n,k+1}X$.  The same argument shows that the middle term is equal to $M_{n,k+1}Y$.
