Truncated octahedron tiles 3D space. Proof?

Question

where can I find a proof that truncated octahedron tiles Euclidean 3D space?

Answer 1

To tile space, center them at lattice points and at the centers of lattice cubes (i.e. lattice points plus $(\frac{1}{2}, \frac{1}{2}, \frac{1}{2})$). The eight faces of the octahedron face the neighboring 8 vertices of the "other" lattice (the one relatively shifted by $(\frac{1}{2}, \frac{1}{2}, \frac{1}{2})$), while the six truncated corners face the six neighbor vertices on the "same" lattice.

Most sources (often crystallographic) consider this to be obvious enough that it needs no proof.