Finally, let's try to prove that any honeycomb involving twip must be an infinite stack of copies of a single layer which is the prism of a 2D tiling.

Since twap is ruled out, a single twip must be part of an infinite stack of twips. We saw in 'Two prisms' that a twip must be surrounded by other prisms (with parallel axes); so at least any prism directly touching the twip is also part of an infinite stack of prisms. A vertical edge of the twip has a configuration corresponding to one in 2D:

12.12.3

12.6.4

12.4.3.3

12.3.4.3

A 12-gon or 6-gon can be converted to 3-gons and 4-gons (as described in the first post here), without changing the validity of any 2D tiling it's a part of. Similarly, in 3D, an infinite stack of twips or hips can be converted to a stack of trips and cubes, without changing the validity of any honeycomb it's a part of. Let's do so, whenever possible.

In 2D, the 12-gon produces 4.3.4.3.3 vertices inside of it (and one 3.3.3.3.3.3 vertex), and incomplete 4.3 vertices at the boundary.

Suppose there's an infinite stack of cubes attached to an infinite stack of trips (with both stacks extending in the same direction, of course). Notice that the analysis of 'Two prisms' was mostly local, considering only a neighbourhood of a vertex. It doesn't matter whether there are other trips and cubes forming a complete twip; all that matters is this configuration {4} 180° {4} 150° {4} 180° {4} 150°. It was shown in 'Two prisms' that these squares must attach to other prisms with parallel axes. Therefore, the two stacks of trips and cubes must attach to other stacks of prisms (except possibly on the far side of the cube, opposite the trip).

Suppose there are three infinite stacks of cubes, corresponding to the 2D configuration 4.4.4 . For the partial sum 3*90°, the program gives us 4*90°, and 3*90° + 2*45°. The latter sum can't be used, as J4 would give an adjacent edge 2*90° + 144.7356° which can't be completed. The former sum can only mean another cube (since the other 90° dihedral angles in elementary polyhedra are not 4.4 as required). So we get a fourth infinite stack of cubes, corresponding to the 2D configuration 4.4.4.4 .

Suppose there are four infinite stacks of trips, corresponding to the 2D configuration 3.3.3.3 . For the partial sum 4*60°, the program gives us 6*60°, and 4*60° + 120°. The dihedral angle 60° appears only in trip, and 120° appears only in hip. And a triangle in trip (in one layer of the honeycomb) can't attach to a hexagon in hip (in the next layer). So we get either two more infinite stacks of trips, corresponding to the 2D configuration 3.3.3.3.3.3, or an infinite stack of hips, corresponding to 3.3.3.3.6 (but then the hips decompose into trips anyway).

Now we only need to prove the following: If we're fitting squares and triangles together in 2D, starting with a 4.3.4.3.3 vertex, completing any vertices including the partial configuration 4.3, or 4.4.4 or 3.3.3.3, and never completing a vertex with just 4.4 or 3.3 or 3.3.3, then the entire plane will be covered. And this should be true regardless of what choices are made in completing vertices (that is, whether any 4.3 becomes 4.3.4.3.3 or 4.3.3.4.3 or 4.3.3.3.4).

If we can prove that, then the statement about twip honeycombs will follow. Each time we complete a vertex in 2D, the corresponding honeycomb gets extended, necessarily with infinite stacks of prisms. When the triangles and squares cover the plane, the corresponding stacks of trips and cubes fill the entire space. When the triangles and squares overlap, invalidating the tiling, the corresponding honeycomb is also invalid. When there's a 30° gap in the tiling, there's also a 30° gap in the honeycomb, which can't be filled by any CRF dihedral angle. I make no claims about a twip honeycomb that's incomplete and cannot be completed. My claim is that any complete honeycomb containing twip must be made of infinite stacks of prisms with parallel axes.