Ternary scale theorems: Difference between revisions

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* A.3.iii. Since ''I''''R'' gains a '''W''' and loses an '''X''' relative to ''I'', the lost letter '''X''' is at the leftmost position of I's window, which is ''p''.

* A.3.iv. Conclusion: ''T''[''p''], the leftmost letter of {{nowrap|''I'' {{=}} ''T''[''p'' : ''p'' + ''k''],}} is '''X'''.

* B.1. Now we go back to the original necklace ''w''. Lift each perfect generator window (we have {{nowrap|''n'' − 1}} perfect windows) of ''T'' to ''w''.

* B.1. Now we go back to the original necklace ''s''. Lift each perfect generator window (we have {{nowrap|''n'' − 1}} perfect windows) of ''T'' to ''s''.

* B.2. By the hypothesis that ''w'' has an AGS, and since the AGS descends to stacking a single generator in the template MOS ''T'', the lifted generators ''g''1 and ''g''2 alternate in their counts of '''Y''' and also alternate in their counts of '''Z'''.

* B.2. By the hypothesis that ''s'' has an AGS, and since the AGS descends to stacking a single generator in the template MOS ''T'', the lifted generators ''g''1 and ''g''2 alternate in their counts of '''Y''' and also alternate in their counts of '''Z'''.

* B.3. For a MOS binary word, the count of a given letter in a generator is coprime to the total count of that letter in one period of the MOS. By this fact applied to ''T'', {{nowrap|gcd(''j'', 2''b'') {{=}} 1}}.

* B.4. Hence, since every instance of the generator in ''T'' has ''j''-many '''W''' letters, every instance of ''g''1 and every instance of ''g''2 has ''j''-many non-'''X''' letters.

* C.1. Importantly, deleting '''X''''s gives windows of length ''j'', such that when you project adjacent lifted generators (by deleting '''X''''s) to the binary necklace {{nowrap|''U'' :{{=}} ''E'''''X'''(''w'')('''Y''', '''Z''')}}, the resulting ''j''-step windows in ''U'' are adjacent and do not overlap.

* C.1. Importantly, deleting '''X''''s gives windows of length ''j'', such that when you project adjacent lifted generators (by deleting '''X''''s) to the binary necklace {{nowrap|''U'' :{{=}} ''E'''''X'''(''s'')('''Y''', '''Z''')}}, the resulting ''j''-step windows in ''U'' are adjacent and do not overlap.

* C.2. Moreover, for every ''j''-step window {{nowrap|''U''[''q'' : ''q'' + ''j'']}}, there exists an {{nowrap|(''i'' + ''j'')-step}} window {{nowrap|''w''[''r'' : ''r'' + ''i'' + ''j'']}} such that {{nowrap|''w''[''r'']}} is the non-'''X''' that corresponds to {{nowrap|''U''[''q'']}} under step deletion. Since by subclaim A, the unique imperfect {{nowrap|(''i'' + ''j'')-step}} window in ''w'' begins in an '''X''', we know that {{nowrap|''w''[''r'' : ''r'' + ''i'' + ''j'']}} is perfect.

* C.2. Moreover, for every ''j''-step window {{nowrap|''U''[''q'' : ''q'' + ''j'']}}, there exists an {{nowrap|(''i'' + ''j'')-step}} window {{nowrap|''s''[''r'' : ''r'' + ''i'' + ''j'']}} such that {{nowrap|''s''[''r'']}} is the non-'''X''' that corresponds to {{nowrap|''U''[''q'']}} under step deletion. Since by subclaim A, the unique imperfect {{nowrap|(''i'' + ''j'')-step}} window in ''s'' begins in an '''X''', we know that {{nowrap|''s''[''r'' : ''r'' + ''i'' + ''j'']}} is perfect.

* C.3. Also note that we only need to stack {{nowrap|2''b'' ≤ ''n'' − 1}} generators to witness this alternation. Under the ordering induced by this stacking, the 1st ''j''-step subword of ''U'' and the {{nowrap|2''b''-th}} ''j''-step window differ due to parity. Since {{nowrap|gcd(''j'', 2''b'') {{=}} 1}}, this visits every note of ''U''.

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 * A.3.iii. Since ''I''<sub>''R''</sub> gains a '''W''' and loses an '''X''' relative to ''I'', the lost letter '''X''' is at the leftmost position of <i>I</i>'s window, which is ''p''.
 * A.3.iv. Conclusion: ''T''[''p''], the leftmost letter of {{nowrap|''I'' {{=}} ''T''[''p'' : ''p'' + ''k''],}} is '''X'''.
-* B.1. Now we go back to the original necklace ''w''. Lift each perfect generator window (we have {{nowrap|''n'' &minus; 1}} perfect windows) of ''T'' to ''w''.
+* B.1. Now we go back to the original necklace ''s''. Lift each perfect generator window (we have {{nowrap|''n'' &minus; 1}} perfect windows) of ''T'' to ''s''.
-* B.2. By the hypothesis that ''w'' has an AGS, and since the AGS descends to stacking a single generator in the template MOS ''T'', the lifted generators ''g''<sub>1</sub> and ''g''<sub>2</sub> alternate in their counts of '''Y''' and also alternate in their counts of '''Z'''.
+* B.2. By the hypothesis that ''s'' has an AGS, and since the AGS descends to stacking a single generator in the template MOS ''T'', the lifted generators ''g''<sub>1</sub> and ''g''<sub>2</sub> alternate in their counts of '''Y''' and also alternate in their counts of '''Z'''.
 * B.3. For a MOS binary word, the count of a given letter in a generator is coprime to the total count of that letter in one period of the MOS. By this fact applied to ''T'', {{nowrap|gcd(''j'', 2''b'') {{=}} 1}}.
 * B.4. Hence, since every instance of the generator in ''T'' has ''j''-many '''W''' letters, every instance of ''g''<sub>1</sub> and every instance of ''g''<sub>2</sub> has ''j''-many non-'''X''' letters.
-* C.1. Importantly, deleting '''X''''s gives windows of length ''j'', such that when you project adjacent lifted generators (by deleting '''X''''s) to the binary necklace {{nowrap|''U'' :{{=}} ''E''<sub>'''X'''</sub>(''w'')('''Y''', '''Z''')}}, the resulting ''j''-step windows in ''U'' are adjacent and do not overlap.
+* C.1. Importantly, deleting '''X''''s gives windows of length ''j'', such that when you project adjacent lifted generators (by deleting '''X''''s) to the binary necklace {{nowrap|''U'' :{{=}} ''E''<sub>'''X'''</sub>(''s'')('''Y''', '''Z''')}}, the resulting ''j''-step windows in ''U'' are adjacent and do not overlap.
-* C.2. Moreover, for every ''j''-step window {{nowrap|''U''[''q'' : ''q'' + ''j'']}}, there exists an {{nowrap|(''i'' + ''j'')-step}} window {{nowrap|''w''[''r'' : ''r'' + ''i'' + ''j'']}} such that {{nowrap|''w''[''r'']}} is the non-'''X''' that corresponds to {{nowrap|''U''[''q'']}} under step deletion. Since by subclaim A, the unique imperfect {{nowrap|(''i'' + ''j'')-step}} window in ''w'' begins in an '''X''', we know that {{nowrap|''w''[''r'' : ''r'' + ''i'' + ''j'']}} is perfect.
+* C.2. Moreover, for every ''j''-step window {{nowrap|''U''[''q'' : ''q'' + ''j'']}}, there exists an {{nowrap|(''i'' + ''j'')-step}} window {{nowrap|''s''[''r'' : ''r'' + ''i'' + ''j'']}} such that {{nowrap|''s''[''r'']}} is the non-'''X''' that corresponds to {{nowrap|''U''[''q'']}} under step deletion. Since by subclaim A, the unique imperfect {{nowrap|(''i'' + ''j'')-step}} window in ''s'' begins in an '''X''', we know that {{nowrap|''s''[''r'' : ''r'' + ''i'' + ''j'']}} is perfect.
 * C.3. Also note that we only need to stack {{nowrap|2''b'' &le; ''n'' &minus; 1}} generators to witness this alternation. Under the ordering induced by this stacking, the 1st ''j''-step subword of ''U'' and the {{nowrap|2''b''-th}} ''j''-step window differ due to parity. Since {{nowrap|gcd(''j'', 2''b'') {{=}} 1}}, this visits every note of ''U''.