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. This contradicts the fact that W₁(λ, σ) and W₂(λ, σ) ~~has~~ different numbers of λ's and σ's.

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. This contradicts the fact that W₁(λ, σ) and W₂(λ, σ) have different numbers of λ's and σ's.

=== Preservation of generators ===

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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. This contradicts the fact that W₁(λ, σ) and W₂(λ, σ) has different numbers of λ's and σ's.
+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. This contradicts the fact that W₁(λ, σ) and W₂(λ, σ) have different numbers of λ's and σ's.
 === Preservation of generators ===