Partition of Facets of Rubik's Cube

Theorem
Let $S$ denote the set of the facets of Rukik's cube.

Then $S$ can be partitioned as follows:
 * $S = \set {S_C \mid S_E \mid S_Z}$

where:
 * $S_C$ denotes the set of corner facets
 * $S_E$ denotes the set of edge facets
 * $S_Z$ denotes the set of center facets.

Proof
From the definition of the facets, each face is divided into $9$ facets.


 * RubiksCubeFacets.png

A facet is either:
 * on the corner of a face, for example $flu$, $fru$
 * on the edge of a face, for example $fu$, $fr$
 * in the center of a face, for example $F$.


 * $(1):\quad$ Each facet can be either in $S_C$ or $S_E$ or $S_Z$ and can not be in more than one.
 * $(2):\quad$ Each facet can be either in $S_C$ or $S_E$ or $S_Z$ and there are no other possibilities.
 * $(3):\quad$ None of $S_C$, $S_E$ and $S_Z$ is empty.

Thus the criteria for $S = \set {S_C \mid S_E \mid S_Z}$ to be a partition are fulfilled.