Recursive structure of MOS scales: Difference between revisions

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Note that the latter two words have at most k s's, and that W₂(Lr+1s, Lrs) has k chunks in total.

~~If w₂ contains fewer complete chunks (<L...Ls preceded by where < is the left chunk boundary) or at least 2 more than w₃, we are done, since they must automatically have different numbers of s's.~~ It suffices to consider the case where the intersection w₂ ∩ w₃ contains at least k-1 complete chunks, since otherwise W₂(Lr+1s, Lrs) would have at least k+1 chunks, which would contradict the length of W₂(λ, σ). 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(Lr+1s, Lrs) 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).

It suffices to consider the case where the intersection w₂ ∩ w₃ contains at least k-1 complete chunks, since otherwise W₂(Lr+1s, Lrs) would have at least k+1 chunks, which would contradict the length of W₂(λ, σ) (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(Lr+1s, Lrs) 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).

[Something's still off, as it doesn't seem to use "reduced word is a non-mos arrangement of two chunk sizes" ⇒ "w is not a mos"...]

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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.
-If w₂ contains fewer complete chunks (<L...Ls preceded by where < is the left chunk boundary) or at least 2 more than w₃, we are done, since they must automatically have different numbers of s's. It suffices to consider the case where the intersection w₂ ∩ w₃ contains at least k-1 complete chunks, since otherwise W₂(L<sup>r+1</sup>s, L<sup>r</sup>s) would have at least k+1 chunks, which would contradict the length of W₂(λ, σ). 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).
+It suffices to consider the case where the intersection w₂ ∩ w₃ contains at least k-1 complete chunks, since otherwise W₂(L<sup>r+1</sup>s, L<sup>r</sup>s) would have at least k+1 chunks, which would contradict the length of W₂(λ, σ) (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).
 [Something's still off, as it doesn't seem to use "reduced word is a non-mos arrangement of two chunk sizes" ⇒ "w is not a mos"...]