Rank-3 scale theorems: Difference between revisions

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 This in particular implies that xyzyx is not LQ.
 ====== MV3 + EMOS implies PMOS (WIP) ======
+Suppose w_{YZ} is a word made by identifying Y and Z into one letter, say Q.
+Suppose w_{YZ} has a k-step, k > 1, which comes in three sizes
+* v_1 = A_1 X + B_1 Q,
+* v_2 = A_2 X + B_2 Q,
+* v_3 = A_3 X + B_3 Q.
+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
+* v_1 = AX + BQ,
+* v_2 = (A-1)X + (B+1)Q,
+* v_3 = (A+1)X + (B-1)Q.
+Eliminating X from w, we have that w_X 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.
+Some B_0-step, say u_L, occurs at least K = ceil(n(w_X)/2) = ceil((b+c)/2) times. If K = 1, then b = c = 1, this case should be relatively easy to deal with...
+If K >= 2, then ...
+Chunks of consecutive X's, separated by at most 2 non-X letters, can come in sizes that differ by at most 2 (including 0), by MV3.
 ====== PMOS implies AG (except in the case xyxzxyx) (WIP) ======