This is coming to you from Yiwei LI (PhD, Applied math), Taiyuan University of Science and Technology  (TYUST) Taiyuan, China

It's going on here for the third round of learning of Birkar's BAB-paper (v2), with scenarios of chess stories. No profession implications. 

Instead of getting "interested" in any thing, I prefer to look into the mechanisms♪ behind getting "interested".

Th 2.15    Th 1.8            ♖    ♘

     ↓   ↖     ↓

Th 1.1      Th 1.6            ♔    ♗

    Mathematics vs Palace stories.(v2)

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Note: technical theorem is not on the board.

♙ ℂ ℍ ℕ ℙ ℚ ℝ ℤ ℭ ℜ I|φ∪∩∈ ⊆ ⊂ ⊇ ⊃ ⊄ ⊅ ≤ ≥ Γ Θ α Δ δ μ ≠ ⌊ ⌋ ∨∧∞Φ⁺⁻⁰ 1

(continued) From the end of move five, one sees a strong implication to calculate (L' + P')|S'. On the other hand, as one sees on the path of thought, the third graph, L' + P' is going to meet B⁺s. This meeting has to be realized by the adjuction on S (in the space of X'), as B⁺s is prescribed in the realm of S. The overall goal is to construct B⁺, yet it is counted on B⁺s. It would be convenient to get B⁺ from B⁺s directly —— though not the case, this thought might help one think of the KV vanishing theorem. Another point is that, B⁺s is only available on the nominal level. That is to say, in the view of construction, B⁺s is just a design on the blueprint, a guidance to the practice. Again, it appears to me, the KV vanishing theorem has a strong (and general) implication concerning the construction of a "plt type" boundary (one may want to review the vanishing theorem) —— here L' as a consequence of some tentative trials (as I guess), then P' as an aide to form the desired "plt type" boundary which happens to take a form of L' + P'. An instructive point in mathematics is that, if one is on the right path, coincidences are awaiting somewhere ahead. The task is to find out the connections among the four roles ——

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 (L' + P')|S'     B⁺s          

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  L' + P'           B⁺

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Before the description of move six, I suddenly think of my bachelor era. I entered a reasonable university free of the national entrance examination for universities before I finished the last term in high school. At that time, around 7% high school students entered universities at last. There was a selection examination for 15 candidates recommended from several notable high schools in the province where I was given birth. Three were recommended from my high school, one boy (me) and two girls, among ~200 classmates. We were top students in a sense but not No. 1 in the classes or in the province, of course. There was an interview for the final decision. I showed to the two officers the certificates that I earned from the national competitions on physics (class/ 2) and math (class/ excellence). I still keep the certificates. Upon the notification, however, I was allocated to the department of chemistry. Only at a later time, I was notified for the second time to participate the department of mathematics. Before the opening of the new campus life, I had three idle months. I bought a book titled "Discrete Mathematics" for self-learning. University textbooks were rarely sold in bookstores at that time (1993), at a capital^. I still keep the book, compiled by Prof. Su-Mei LI, printed by Tsinghua Univ. Press. I finished learning one or two chapters at home. In retrospect, this good start was interrupted after entering the university... The situation was somewhat like —— you wanted to be a "painter", yet for years you were arranged to measure the sizes of the painting brushes, or bother about the physics of pigments, that had nothing to do with becoming a "painter".

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Return to the first sentence of this note, one has ——

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(L' + P')|S' = L'|s' + P'|s'.

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If one still remembers the expression L' = c'n = n'c + d'n = 2M' + n·(M' - (Kx' + B')) + nΔ' - ⌊(n + 1)Δ'⌋, one may calculate ——

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2M'|s' ~ 0,

n(M' - (Kx' + B'))|s' = nM'|s' - n(Kx' + B')|s' ~ - n(Kx' + B')|s' defined as -n(Ks' + Bs'),

nΔ'|s' is defined as nΔs',

⌊(n + 1)Δ'⌋|s' = ⌊(n + 1)Δ'|s'⌋ = ⌊(n + 1)Δs'⌋.

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All together, one has L'|s' ~ -n(Ks' + Bs') + nΔs' - ⌊(n + 1)Δs'⌋. Now, define Ps':= P'|s', one has ——

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(L' + P')|s' ~ -n(Ks' + Bs') + nΔs' - ⌊(n + 1)Δs'⌋ + Ps'.    (%)

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So far, one has not seen the connection between (L' + P')|s' and B⁺s. Observe that Bs' and  B⁺s are somewhat close by appearances. In particular, one can view Bs as part of B⁺s, defining Rs:= B⁺s - Bs which is non-negative (why introduce Rs ?). In view of X'-space, one has Rs' = B⁺s' - Bs' from which one "solves" for Bs' = B⁺s' - Rs'. By applying the definition (2nd item) of n-complement, one has -n(Ks' + Bs') = - n(Ks' + B⁺s' - Rs') = -n(Ks' + B⁺s') + nRs' ~ nRs'. More explicitly, n(Ks' + B⁺s') ~ 0 is applied as a property of n-complement in the X'-space. Now, the expression (%) takes a form of ——

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(L' + P')|s' ~ nRs' + nΔs' - ⌊(n + 1)Δs'⌋ + Ps'.

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The right-hand side is assigned a notation of Gs'. (To look ahead, one may view Gs' in X-space: Gs = nRs, or Rs = Gs/n, as nΔ = ⌊(n + 1)Δ⌋ and P' exceptional/ X). The introduction of Rs (or Rs') is of theory of mathematics (TOM): remove any redundant ingredients before moving forward. Rs is the essential ingredient contained in B⁺s, of which Bs is not new, as the outputs of Th 2.13 customized to the adjunction on S of the main pair (X, B).

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The demand for proving Gs' non-negative appears to imply that it is not clear if L' is non-negative upon definition. Also, the information of L' + P' ~ G' ≥ 0 only comes from (L' + P')|s' ~ Gs' ≥ 0. It is not clear if L' + P' ≥ 0 or (L' + P')|s' ≥ 0 hold (seemingly not used, but better to know). It appears to me that, L' + P' ~ G' ≥ 0 or (L' + P')|s' ~ Gs' ≥ 0 are not required by the KV vanishing theorem. These non-negative requirements are originated from their roles in forming the non-negative increment in B⁺. (To look ahead, B⁺ = B + R with R:= G/n, while the meaning of G comes from L ~ G ≥ 0).

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In summary, the raw thought of move six is just to calculate (L' + P')|s' and have the outcome Gs' related to B⁺s through its essential ingredient Rs in X'-space.

ℭ ℜ I|φ∪∩∈ ⊆ ⊂ ⊇ ⊃ ⊄ ⊅ ≤ ≥ Γ Θ α Δ δ μ ≠ ⌊ ⌋ ⌈ ⌉ ∨∧∞Φ⁺⁻⁰ 1

Calling graph for the technical theorem (Th1.9) ——

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Th1.9

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[5, 2.13(7)]   Lem 2.26   Pro4.1   Lem2.7

                                                          |

.....................................................Lem2.3   

Note: Th1.9 is only called by Pro.5.11, one of the two devices for Th1.8, the executing theorem.

Pro4.1                                                    

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[5, ?]   [37, Pro3.8]   [5, Lem3.3]   Th2.13[5, Th1.7]   [16, Pro2.1.2]  [20]  [25, Th17.4]

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Special note: Original synthesized scenarios in Chinese for the whole proof of v1 Th1.7, the technical theorem.

*It's now largely revised* due to new understandings.

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See also: Earlier comments in Chinese* (v1).

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It is my hope that this action would not be viewed from the usual perspective that many adults tend to hold.