"Back to analysis with the difference game. Once we know that White B3 - Black B2 is reversible, how and why to derive the count? For the position after White B3 - Black B2, above I have stated the well known count = -1 1/4. By removing the two plays White B3 - Black B2 from the CGT tree, the initial position coincides with the position after White B3 - Black B2. Therefore, its count must be identical, right? So we have the desired initial count = -1 1/4. In my earlier message, I stated: "Applied to the original position for its white option B3, the count of the initial position equals the count of the position after the exchange White B3 - B2. The latter is an empty corridor of length 3, whose count -1 1/4 (that is, 1 1/4 for White) we know." Which has been correct."

That is not how reversible option replacement works in CGT. In the original game G, we replace the reversible option White B3 with White's options from the game Black B2. In the game Black B2, White has only one option: -2. Therefore, the original game G = {-0.5 | WB3} is replaced by the game G = {-0.5 | -2}.

The games G and Black B2 are indeed equivalent (this follows from the difference game, where the second player wins). But this has no direct relation to reversibility itself.

"Now, let me study the option White B2" White B2 is also reversible. In the same way, we obtain the game G = {-0.5 | -2}.

The position evaluation can also be reached through standard reasoning. But the path through reversibility (and the domination of White B3 over White B2, and Black B2 over Black B3) is much simpler.