"It is not true that if t(C) = t(A) that A reverses through C": Indeed. I think I have found a few examples rejecting this. Playing the difference game yielded different results than comparing move values.
A few weeks ago, I have proven the proposition below and its analogue for White's 3-move sequence. In particular, it relates C ≤ A <=> MA ≤ MC, as we want. Furthermore, a hundred examples exhibit this relation. Some of the examples have more complicated trees with unsettled PE, an alternating gote or sente sequence for White's start, PB or PE as simple sente or long sequences resulting in series of such conditions. I do not have any counter-example.
We want C ≤ A <=> MA ≤ MC for the general case, but so far we cannot prove it. The proposition presumes settled PD and PE and gote traversal values of PA. Obviously, the requirements of PE can be relaxed to generalise the proposition with more research effort. E.g., we can allow privileges. However, real problem is: what is the most general case?
You have wanted to claim C ≤ A <=> MA ≤ MC in the general case, but you need to prove it. Do my proposition and proof below give you enough information for this purpose?
If other methods, such as comparing the opponent's branches or playing the difference game, are inapplicable, you have suggested MA ≤ MC as the always applicable condition for identifying or rejecting traversal of a 3-move sequence worth playing successively. Prove it! :) Otherwise, I ask again: how, in the general case, do we distinguish possible from impossible reversal (to start with, at least for alternating 3-move sequences)?
For proposition 10, the game tree represented by the counts is:
Code:
.....A...
..../.\..
...B...E.
../.\....
.F...C...
..../.\..
...D...G.
An 'unsettled' position corresponds to a combinatorial game that is not a number. Initially, the condition A = (D + E) / 2 is the tentative gote traversal count of PA. The theorem confirms that we have this gote traversal count and Black's 3-move traversal sequence because the conditions C ≤ A < D <=> C ≤ (D + E) / 2 < D in proposition 10 express that Black's alternating sequence is worth playing successively, that is, can be traversed. During Black's initial sente sequence, the count A of the initial position becomes the smaller or equal count C so Black can incur an initial loss. He compensates it by moving to the count D of his final gote follower because D is larger than A, which means his gain.
We have Black's alternating 3-move sequence PA - PB - PC.
Presuppositions
0) We have the combinatorial game PA := {PB|PE} := {PF|PC||PE} := {PF||PD|PG|||PE} with unsettled PA, PB, PC and settled PD, PE.
1) A := (D + E) / 2.
2) MA := (D - E) / 2.
3) C := (D + G) / 2.
4) MC := (D - G) / 2.
Proposition 10 [Black's 3-move gote]
G ≤ E <=> C ≤ A <=> MA ≤ MC and A < D.
Proof
Part I: "MA ≤ MC" <=> "G ≤ E <=> C ≤ A":
MC ≥ MA <=>(2)(4) (D - G) / 2 ≥ (D - E) / 2 <=> D - G ≥ D - E <=> -G ≥ -E <=> G ≤ E <=> G/2 ≤ E/2 <=> (D + G) / 2 ≤ (D + E) / 2 <=>(1)(3) C ≤ A.
Part II: "A < D":
PA unsettled(0) => D > E => D > (D + E) / 2 > E =>(1) D > A.
Remarks
(1) and (2) say that PA has the count A and move value MA so is Black's long gote with his 3-move gote sequence. (3) and (4) characterise PC as a simple gote. Note that MA = MC is allowed but we need not speak of it as an 'ambiguous' case.
When applying proposition 10 to an example, we check the presuppositions (0), (3) and (4) and whether therefore PC is a simple gote indeed, make the hypothesis of the count A, need not verify G ≤ E, verify C ≤ A to confirm the count A and calculate the move value MA. If the sequence to PD follows dominating options, we need not verify A < D because it follows from the already checked presuppositions (0) and (1). However, if we do not know whether the sequence to PD follows dominating options, we can determine this by verifying or refuting A < D.
Informally, we call A and MA in the presuppositions tentative values. The proposition confirms the values so, after verification of the conditions, we can remove the 'tentative' tag. The proposition and proof imply that the count A and move value MA of PA are well-defined.
Future research can make proposition 10 more flexible for PE if only it does not alter (1) and (2).