Rubik's Cube Stage 6 -- show bottom two layers are preserved by $ R^{-1}FR^{-1}BBRF^{-1}R^{-1}BBRRU^{-1} $ Rubik's cube blog has for stage 6 a sequence of 13 moves to be iterated: $$ R^{-1}FR^{-1}BBRF^{-1}R^{-1}BBRRU^{-1} $$
I've noticed as we progress through the Rubik's cube algorithm two things happen:


*

*we conserve more and more cubes (pieces stay where they are)

*the sequences get longer and more complex


Here I just want to see a proof that this sequence works - that it preserves the bottom two layers.  Even if someone gives a presentation and punches these sequence of moves in a calculator.

Lastly, since only 9 moves are being permuted, can we write down the explicit permutation this corresponds to?
 A: The mathematical proof THAT the sequence works would be:

Proof. By direct computation.

which assumes that the skeptical reader will know how to compute compositions of permutations himself and see that the composition of all the moves is a permutation that indeed leaves the lower two layers unchanged.
The physical cube is a specialized mechanical tool for carrying out such computations in the cube group, but it is not really any less mathematical than it is to use the mechanical tools known as "pencil and paper" to keep track of the same computation. Or an electronic tool: A computer.
A pencil-and-paper computation could take the form of a large table detailing where each cubie (or each sticker) is at each step in the operation. It is fairly easy (yet too tedious for me to bother) to write down this table with the help of a physical cube. Verifying the table symbolically is even more tedious (you need to write down the action of each basic turn and check that it describes the difference between each line and the next), but not particularly difficult either.

HOW it works: We can decompose the sequence as
$$ \underbrace{R^2\underbrace{RFR'B^2(RFR')'B^2}_{[RFR',B^2]}R^2}_{\text{conjugated by }R^2}U' $$
The core of this operation is the commutator $[RFR',B^2]=RFR'B^2RF'R'B^2$, which permutes the three corner positions (bul), (bdr), (fdr) cyclically and leaves everything else unchanged. Like many 3-cycles, it is constructed from two simpler operations $RFR'$ and $B^2$ whose effects intersect at exactly one cubie, namely the (bdr) corner. A commutator of two such combinators will cycle the shared point and the two cubies that its parts move into that point.
This is not the shortest way to cycle three corners, but has been picked because it preserves which side of the three corners point UP/DOWN.
The commutator is then conjugated with $R^2$, turning it into a 3-cycle of (bul), (fur), (bur) which preserves the top colors.
Finally a single $U'$ at the end moves 2 of the top corners back to their original positions -- with the net effect that (ful) and (fur) are now swapped -- at the cost of (only now!) disrupting the top edges.
A: here is the result of the top face on a solved cube (up to rotation)

