Ternary scale theorems: Difference between revisions

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'''Claim 1''': Deleting X's from the alternant subwords of ''S'' gives every ''j''-step subword in the scale ''E''<sub>X</sub>(''S'')(Y, Z), the scale word obtained by deleting all X's from ''S''.

Proof: Consider the "imperfect alternant" of ''S'' on index ''p'', which is ''I'' = ''S''[''p'' : ''p'' + ''i'' + ''j''], and suppose the result of deleting all X's from ''I'' has a j-step subword ''w'' as a substring. Then ''S''[''p'' − 1: ''p'' − 1 + ''i'' + ''j''] and ''S''[''p'' + 1 : ''p'' + 1 + ''i'' + ''j''] are both detemperings of perfect generators of ''T'', and have one fewer step ~~that is Y or Z~~ by our assumption. Thus the word ''I'' must both begin and end in a non-X letter. Removing all the X's from ''I'' results in a word that is ''j'' + 1 letters long and is the ''j''-step word ''w'' we started with, with just one extra letter appended. Thus one of the two perfect generators above, namely the one that removes the extra letter, must contain this ''j''-step. The rest of the ''j''-step subwords are all contained in "perfect" alternants; take ''q'' ≠ ''p'' to be the index of the first letter of one such ''j''-step subword (as contained in ''S'') and use S[''q'' : ''q'' + ''i'' + ''j''].

Proof: Consider the "imperfect alternant" of ''S'' on index ''p'', which is ''I'' = ''S''[''p'' : ''p'' + ''i'' + ''j''], and suppose the result of deleting all X's from ''I'' has a j-step subword ''w'' as a substring. Then ''S''[''p'' − 1: ''p'' − 1 + ''i'' + ''j''] and ''S''[''p'' + 1 : ''p'' + 1 + ''i'' + ''j''] are both detemperings of perfect generators of ''T'', and have one fewer non-X step than ''I'' by our assumption. Thus the word ''I'' must both begin and end in a non-X letter. Removing all the X's from ''I'' results in a word that is ''j'' + 1 letters long and is the ''j''-step word ''w'' we started with, with just one extra letter appended. Thus one of the two perfect generators above, namely the one that removes the extra letter, must contain this ''j''-step. The rest of the ''j''-step subwords are all contained in "perfect" alternants; take ''q'' ≠ ''p'' to be the index of the first letter of one such ''j''-step subword (as contained in ''S'') and use S[''q'' : ''q'' + ''i'' + ''j''].

'''Claim 2''': If a binary scale ''U'' has ''b'' Y's and ''b'' Z's, gcd(''j'', 2''b'') = 1, and consecutively stacked ''j''-steps in ''U'' occur in 2 alternating sizes, then ''U'' = (YZ)<sup>''b''</sup>.

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 '''Claim 1''': Deleting X's from the alternant subwords of ''S'' gives every ''j''-step subword in the scale ''E''<sub>X</sub>(''S'')(Y, Z), the scale word obtained by deleting all X's from ''S''.
-Proof: Consider the "imperfect alternant" of ''S'' on index ''p'', which is ''I'' = ''S''[''p'' : ''p'' + ''i'' + ''j''], and suppose the result of deleting all X's from ''I'' has a j-step subword ''w'' as a substring. Then ''S''[''p'' &minus; 1: ''p'' &minus; 1 + ''i'' + ''j''] and ''S''[''p'' + 1 : ''p'' + 1 + ''i'' + ''j''] are both detemperings of perfect generators of ''T'', and have one fewer step that is Y or Z by our assumption. Thus the word ''I'' must both begin and end in a non-X letter. Removing all the X's from ''I'' results in a word that is ''j'' + 1 letters long and is the ''j''-step word ''w'' we started with, with just one extra letter appended. Thus one of the two perfect generators above, namely the one that removes the extra letter, must contain this ''j''-step. The rest of the ''j''-step subwords are all contained in "perfect" alternants; take ''q'' &ne; ''p'' to be the index of the first letter of one such ''j''-step subword (as contained in ''S'') and use S[''q'' : ''q'' + ''i'' + ''j''].
+Proof: Consider the "imperfect alternant" of ''S'' on index ''p'', which is ''I'' = ''S''[''p'' : ''p'' + ''i'' + ''j''], and suppose the result of deleting all X's from ''I'' has a j-step subword ''w'' as a substring. Then ''S''[''p'' &minus; 1: ''p'' &minus; 1 + ''i'' + ''j''] and ''S''[''p'' + 1 : ''p'' + 1 + ''i'' + ''j''] are both detemperings of perfect generators of ''T'', and have one fewer non-X step than ''I'' by our assumption. Thus the word ''I'' must both begin and end in a non-X letter. Removing all the X's from ''I'' results in a word that is ''j'' + 1 letters long and is the ''j''-step word ''w'' we started with, with just one extra letter appended. Thus one of the two perfect generators above, namely the one that removes the extra letter, must contain this ''j''-step. The rest of the ''j''-step subwords are all contained in "perfect" alternants; take ''q'' &ne; ''p'' to be the index of the first letter of one such ''j''-step subword (as contained in ''S'') and use S[''q'' : ''q'' + ''i'' + ''j''].
 '''Claim 2''': If a binary scale ''U'' has ''b'' Y's and ''b'' Z's, gcd(''j'', 2''b'') = 1, and consecutively stacked ''j''-steps in ''U'' occur in 2 alternating sizes, then ''U'' = (YZ)<sup>''b''</sup>.