User blog comment:Koteitan/Purely mathematical definition of BMS/@comment-30754445-20181123092422/@comment-30754445-20181123103610

Well... almost. The two versions would be exactly equivalent, if I changed my the "strictly smaller" in step 3 to "smaller or equal". And I was correct in assuming that it isn't too difficult to prove.

Here is a general outline of the argument (for the sake of readability, whenever I say "my version" this refers to the corrected version):

1. If a "bad root candidate" Y doesn't obey the constraint in my version, then there's some pair Z between Y and X such that Z0 <= Y0. Either Z is a P0-ancestor of X or it is not.

(a)  If Z is a P0-ancestor of X, then Y (which comes before Z) can only be a P0-ancestor of X if Y is a P0-ancestor of Z. But this is Impossible, since Z0 <= Y0. Ergo, Y is not a P0-ancestor of X.

(b) if Z is not a P0-ancestor of X then this means that any node T to the left of Z for which T0 > Z0cannot be a P0-ancestor of X (pause for moment to think why this is true). Y is such node. Ergo, Y is not a P0-ancestor of X.

So regardless of the status of Z, we've shown that Y is not a P0-ancestor of X. Therefore Y does not obey the constraint of Kotetian's version.

2. if a bad root candidate Y does obey the conTstraint in my version, then there's no Z between Y and X such that Z0<=Y0. It is quite clear that given this, any P0-branch going backwards from X will contain Y (again, pause for a moment to think why this must be true). Therefore Z is a P0-ancestor of X and we're done.

The only question that still remains is this:

Is Koteitan's definition equivalent to my uncorrected version for standard sequences (that is: for sequences that can be arrived at from (0,0,0)(1,1,1) in either version?).

I conjecture that the answer is yes. Though it seems that actually demonstrating this is going to be quite difficult.

(if the answer is "no", however, it could be quite easy to find counterexamples)