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By Thomas Lam, Luc Lapointe, Jennifer Morse, Mark Shimozono

Quantity 208, quantity 977 (second of 6 numbers).

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18, either tq,p y(i) = y(i) or tq,p y(i) ≤ p. 37) we have tq,p y(i) ≤ p0 as desired. 20. In Case C, S is a strong strip. Proof. We use the following notation: S: ··· S1 : ··· S : ··· C −− − a− 1 ,b1 − a− 1 ,b1 C −− G· i,j − Gy i,b1 Gv Gy a− ,b− Gv C− C − Gw i,j C i,b1 C Gu Gx where (a− , b− ) = (y −1 (q), y −1 (p)). 19. By induction and the definition of S , we need only check its last two pairs of consecutive marks. 37). 47) p > y(a− 1 ). − −1 (p0 ) > l and a− Suppose that y(a− 1 ) ≤ p0 . Since y 1 ≤ l we have y(a1 ) < p0 .

Now p < qC are consecutive A∨ -nice integers. Since u(b) is also A∨ -nice it follows that u(b) ≤ p. Suppose u(b) = p. 15 on the non-commuting A i1 ,j1 initial pair (u x, u −→ z), we have the contradiction l < b = u−1 (u(b)) = u−1 (p) = z −1 (p) ≤ l. So u(b) < p. But then condition B is satisfied by the integer p which is less than qC , meaning that Case RB holds, which is a contradiction. 15. In Case RC, u (i1 ) > u (j1 ). Proof. 15) p > tp,qC u(i1 ). 12 we have u(i1 ) < u(b) ≤ p < qC = u(j1 ). 15) clearly holds.

5. (i) In Case RA, m(C) = c−1 A (m(C )). In Case RB, m(C) > m(C ). In Case (m(C )). RC, m(C) ≥ c−1 ∨ A (ii) m(C) > x(b). (iii) Case RC cannot be preceded by Case RB. (iv) The initial pair (W, C) commutes in Cases RA and RC and does not in Case RB. 42 6. 1. Case RA. 14 (w pair such that the diagram commutes. 7 S is a strong strip. 3) as follows. 6. 5 holds. Proof. (iv) was proved above. For (i) we have m(C) = u(a) = c−1 A (x(a)) = a,b a,b c−1 A (m(C )). For (ii) we need to show x(b) < u(a). Since v −→ x and w −→ u are strong covers, we have x(b) < x(a) and u(b) < u(a).

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