Rank-3 scale theorems: Difference between revisions

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If some A_i and A_j differed by more than 2, a contradiction would result, as all intermediate combinations must be attained (since scooting over by one step changes the numbers of Q's and X's by <= 1). So we assume we have sizes
If some A_i and A_j differed by more than 2, a contradiction would result, as all intermediate combinations must be attained (since scooting over by one step changes the numbers of Q's and X's by <= 1). So we assume we have sizes
* v_1 = AX + BQ,
* v_1 = v(s_1) = AX + BQ,
* v_2 = (A-1)X + (B+1)Q,
* v_2 = v(s_2) = (A-1)X + (B+1)Q,
* v_3 = (A+1)X + (B-1)Q.
* v_3 = v(s_3) = (A+1)X + (B-1)Q.
Eliminating X from w, we have that w_X (word in Y and Z) is a mos, by the EMOS assumption. At least one of B, B-1, B+1 is not a period multiple of w_X, say B_0, hence a B_0-step of w_X comes in two sizes. For a contradiction we seek two intervals in w:
# u_L + A_0 X, and
# u_s + A_0 X,
where u_L and u_s are the larger and smaller B_0-steps of w_X.


Plan: v_2 and v_3 are sizes that are problematic when they occur together.
Plan: v_2 and v_3 are sizes that are problematic when they occur together.
Let w_X, w_Y, w_Z be mosses that result from eliminating X, Y and Z. MV3 implies that for any possible choices of s_i, EX(s_1), EX(s_2), EX(s_3) each only comes in one possible size as B-, (B+1)- and (B-1)-steps in w_X.


====== PMOS implies AG (except in the case xyxzxyx) (WIP) ======
====== PMOS implies AG (except in the case xyxzxyx) (WIP) ======