Lemma 4:
I claim that on the n x n x n odd degree cube the number of incorrectly flipped innermost edges must be even. Again this is an easy consequence using group theory, however I do not know it well enough to prove this for the n x n x n case, so I will do it simply using the properties of the n x n x n cube.

Proof:
First I will define a coordinate system that will make it possible to see when an edge is correctly or incorrectly flipped. Similar to in lemma 1 I will order the colors in a certain hierarchy, however the edge piece hierarchy must include 6 colors instead of only 2 as for the corners. Let pairs of opposite colors be numbered in a hierarchy in the following order: The highest tier are the colors of the U and D faces, the middle tier are the colors of the F and B faces, and the lowest tier are the colors of the R and L faces.

Now each edge has either a sticker color of the highest tier, the middle tier, or both. Each position in the innermost edge piece orbit has two sides to it, which are taken up by the two stickers on each innermost edge piece. We can see that one of these sides will lie in a higher rank in the hierarchy of colors, since the two stickers are on adjancent faces, and hence not ranked in the same tier. Define the base of that position to be the side that lies in the face that is ranked highest in the hierarchy of colors out of the two sides. For example for the UF, UR, UB, and UL positions, the U side of each of them would be the base of that position. This is because the U face color is ranked highest in the hierarchy of colors, and would therefore be ranked higher than the F and B sides of the UF and UB positions, as well as the R and L sides of the UR and UL positions. Therefore the base of all the U layer innermost edge positions would be the U side of each of those positions.

Now we have to define when the edges are correctly oriented. First we'll do this for the 8 U and D layer edges. An edge that belongs in the U or D layer when the cube is solved that is currently found in any of the 8 positions in the U or D layer is correctly oriented when its highest tiered sticker color lies in the base of the position. An edge that belongs in the U or D layer when the cube is solved that is found in the E layer of the cube is considered to be correctly oriented when the highest tiered color on that sticker does not lie in the base of the position.

Now onto the E layer edges. When an edge that belongs in the E layer of the cube when it is solved is in the E layer it is considered to be correctly oriented when it's highest tired sticker color lies in the base of that position, otherwise it is incorrectly oriented. If an edge that belongs in the E layer when the cube is solved is found in the U or D layer, then it is considered to be correctly oriented when it's highest tier sticker color is not found in the base of that position, otherwise it is considered incorretly oriented.

Now we need to find all the possible turns on the n x n x n cube that affect the innermost edge orbit. We will, as mentioned on the main proof page, not allow the turning of the middlemost slice on an n x n x n cube. However this can be simulated by turning all the other (n-1) slices the opposite direction and then rotating the cube. Therefore since the turning of the middlemost slice is banned as a possibility, the only turns that affect the innermost edge orbit are the 18 outer face turns, F, F', F2, B, B', B2, R, R', R2, L, L', L2, U, U', U2, D, D', D2. As in lemma 1 we can see that rotating the entire cube preserves the orientation of all the innermost edges based on our definition above, since the pairs of opposite colors in each tier of the hierarchy will change since the colors occupying those positions have changed. This is also easily shown by rotating the entire solved cube, clearly all the innermost edges are still all correctly oriented, hence none of the orientations of the edges have been affected.

Since rotating the cube preserves the orientation of the cube, then we need only examine all the possible turns of one face in each opposite pair of faces. This in effect examines for both faces, since we could simply rotate the opposite face into the position of the face we have examined, and the same results would still hold. 1) Let us examine how all the possible turns on the U face affect the orientations of the innermost edges. For each of the three turns we will start from the completely solved state and perform a turn. This is a valid way to examine the cube since this is how we scramble and solve it in real life.

Perform any turn on the U layer. We can see that all of the pieces still have their highest tier sticker color in the base of its position, namely that the U sticker color of each piece is still in the U layer, therefore a U turn does not affect the orientation of any of the 4 edges contained in that layer.

2) Let us now examine how all the possible turn on the F face affect the orientations of the edges. Again examine how the turn affects the solved state.

a) Perform a single quarter turn of the F face, in particular either F or F'. We can see that the U and D layer pieces end up in the E layer, and that the E layer pieces end up in the U and D layers. Notice though that the turn also moves the higest tier sticker color from the base to not in the base and vice versa. For the U and D pieces, an F face quarter turn places the piece in the E layer, as well as places its highest tier sticker color outside the base of the positions it ends up in. Also notice that the E layer pieces end up in the U and D layers, and that the highest tier sticker color of that piece ends up outside the base of its position. Therefore all 4 pieces are still correctly oriented, and any F face quarter turn leaves the orientations of the 4 innermost edge pieces in that layer alone.

b) Perform the double turn F2 on the F face. We can see that the U and D layer pieces remain in either the U or D layer, and that the E layer pieces remain in the E layer. Also notice that the highest tier sticker color of each piece remains in the base of its new positions. Therefore the double turn F2 does not affect the orientations of any of the 4 innermost edge pieces contained in that layer.

3) Let us now examine how a turn on the R face affects the orientations of the innermost edges. Again start from the solved state to make visualizing easier.

a) Perform a single quarter turn of the R face, in particular either R or R'. We can see that the U and D layer pieces end up in the E layer, and that the E layer pieces end up in the U and D layers. Notice though that the turn also places the highest tier sticker color into the base of the new layer. For the U and D pieces, an R face quarter turn places the piece in the E layer, as well as places its highest tier sticker color into the base of the position it ends up in. Also notice that the E layer pieces end up in the U and D layers, and that the highest tier sticker color of that piece ends up in the base of its new position. Therefore all 4 pieces are now incorrectly oriented, and any R face quarter turn flips the orientation of the 4 innermost edge pieces in that layer. Notice that this turn can only flip the orientation of 4 pieces at a time, an even number. Therefore, starting from the solved state we can only ever change the orientation of an even number of pieces, and hence it is impossible to change the orientation of an odd number of pieces, therefore making the number of incorrectly flipped pieces an odd number.

b) Perform the double turn R2 on the R face. We can see that the U and D layer pieces remain in either the U or D layer, and that the E layer pieces remain in the E layer. Also notice that the highest tier sticker color of each piece remains in the base of its new positions. Therefore the double turn R2 does not affect the orientations of any of the 4 innermost edge pieces contained in that layer.

Therefore we have proved that the number of incorrectly flipped pieces must be even.

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