Let $C$ be a category with cofibrations in the sense of (Waldhausen, Algebraic K-Theory of Spaces) and denote by $F_n(C)$ the category with cofibrations consisting of sequences of $n$ cofibrations $A_0 \rightarrowtail A_1 \rightarrowtail \dotsc \rightarrowtail A_n$ in $C$. A cofibration in $F_n(C)$ is a commutative ladder consisting of "lattices". See the attached PDF for a precise definition.

Waldhausen gives a beautiful graphical proof that there is an equivalence $F_n F_m C \cong F_m F_n C$ (Lemma 1.1.5). The reason is, basically, that an object in both categories is given by a rectangular array of cofibrations, such that each square is a lattice, which is a *symmetric* condition. Now, Waldhausen only indicates why this equivalence $F_n F_m \cong F_m F_n$ also preserves cofibrations, and is therefore exact. Namely, he claims that a cofibration in $F_n F_m C$ is a $3$-dimensional diagram satisfying similarly some *symmetric* condition, but he does not name it explicitly. So what is this condition? So basically the question is: When should we call a commutative cube of cofibrations a lattice?

The full question contains many diagrams (also a 3-dimensional one), which cannot be properly displayed with MO-Latex. Therefore I've decided to write the full question, also with some of my attempts so far, as a PDF. I hope that this is not unappropriate for MO.

EDIT: Eva Höning pointed out to me that a cube of fibrations is a lattice if every one of the six squares is a lattice, and the map from the pushout of the cube without its tip to the tip is a cofibration. And it is rather straight forward to see that both $F_n F_m$ and $F_m F_n$ have these cubes as cofibrations. In the first place I was not careful enough with the definitions to see that, but it is really easy.