Recursive structure of MOS scales: Difference between revisions

# w₂(L, s) = the first K+1 letters of W₂(L^r+1s, L^rs) (which must be longer than w₁(L, s), since W₂ had more λ's than W₁)

# w₃(L, s) = the last K+1 letters of W₂(L^r+1s, L^rs)

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. Hence it suffices to consider the following two cases, each split into subcases:

Case 1.1:

This implies that the last chunk is bigger than the first one, a contradiction because w₂ begins in λ.  

Case 1.2:

  w1:  s[something with k s's]

Truncate the strings as follows, to get three distinct K-steps in w:

Case 1.3:

or

Case 1.4:

(both 1.3 and 1.4 imply w₃ has one more s).

Case 2.1:

Case 2.2:

⇒ do the same trick as in Case 1.2

Case 2.3:

Case 2.4:

Both 2.3 and 2.4 are contradictions since the first chunk is λ so has to be at least as big as the last one.