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| === Preservation of the MOS property === | | === Preservation of the MOS property === |
| (Assume that the mos has length n; the notation w(X, Y) for a word w(L, s) in L, s means w with step sizes X substituted for L and Y substituted for s.) | | (Assume that the mos has length ''n''; the notation ''w''(X, Y) for a word ''w''(L, s) in L, s means ''w'' with step sizes X substituted for L and Y substituted for s.) |
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| Suppose w(L, s) had three chunks L...s with r, r+1 and r+2 'L's. Then we have a length r+2 subword that's only 'L's, one that has one s at the end and one that has two 's's on either side, which means that the original scale was not MOS. Therefore the reduced word has two step sizes. | | Suppose w(L, s) had three chunks L...s with ''r'', ''r'' + 1 and ''r'' + 2 'L's. Then we have a length r+2 subword that's only 'L's, one that has one s at the end and one that has two 's's on either side, which means that the original scale was not MOS. Therefore the reduced word has two step sizes. |
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| [Proof is incomplete]<!--
| | We claim that ''T''<sub>s</sub> = the set of indices where s steps occur is a [[Maximal evenness|maximally even]] subset of size |T_s| in '''Z'''/n'''Z'''. |
| Without loss of generality assume r ≥ 1 (otherwise flip the roles of L and s). Let W'(λ, σ) be the reduced word with step sizes λ (corresponding to the chunk of L's of size r+1) and σ (corresponding the chunk of size r), and assume that W' is not a mos. Then for some k, W' must have k-steps of the following sizes:
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| # p₁ λ's and q₁ σ's, represented by subword W₁(λ, σ) in W'.
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| # p₂ λ's and q₂ σ's, represented by subword W₂(λ, σ) in W'. We can assume W₂ begins in λ; otherwise we can slink W₂ to the right until it begins in λ, which is guaranteed to never decrease the number of λ's.
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| Here, pᵢ + qᵢ = k and we assume p₂ - p₁ ≥ 2.
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| Let K = p₁(r + 1) + q₁r + k. Consider the following sizes for (K+1)-steps in w:
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| # w₁(L, s) = the word sW₁(L<sup>r+1</sup>s, L<sup>r</sup>s) [W₁ interpreted as a subword of the original mos], with (k + 1) s's. w₁ is K+1 letters long.
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| # w₂(L, s) = the first K+1 letters of W₂(L<sup>r+1</sup>s, L<sup>r</sup>s) (which must be longer than w₁(L, s), since W₂ had more λ's than W₁)
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| # w₃(L, s) = the last K+1 letters of W₂(L<sup>r+1</sup>s, L<sup>r</sup>s)
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| Note that the latter two words have at most k s's, and that W₂(L<sup>r+1</sup>s, L<sup>r</sup>s) has k chunks in total.
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| It suffices to consider the case where the intersection w₂ ∩ w₃ contains at least k-1 complete chunks, since otherwise we would contradict either the length of W₂(λ, σ) or the mos property of w (Todo: bring back the cases and diagrams to prove this rigorously). If the intersection had exactly k-1 chunks, this implies that one substring is a proper subset of the other, a contradiction. Thus the intersection has to have exactly k chunks, implying w₂ = w₃, and that W₂(L<sup>r+1</sup>s, L<sup>r</sup>s) is exactly K+1 letters long, only one more than w₁(L, s). This contradicts the fact that W₂(λ, σ) has at least two more λ's than W₁(λ, σ).-->
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| === Preservation of generators === | | === Preservation of generators === |