Ternary scale theorems: Difference between revisions

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* In the doubly even case, start from the mode of ''s'' whose template MOS is the brightest mode. Let ''s''2 be offset at a generator of the even-regular scale, which by Theorem 6 we choose to have the same interval class as a bright generator of the MOS ''a'''''x''' 2''k'''''X'''. This is what induces the equality of ''s''1 and ''s''2 (in particular, the two scales have the same period, thus they are both primitive): Let ''s''''t'' be the period of the brightest mode of the template MOS, and let ''g'' be its bright generator class. Then the slice {{nowrap|''s''''t''[-''g'' +1 : 1]}} is the imperfect generator of the MOS. Now when we "darken" the mode by one generator, which is the difference between the template MOSes of ''s''1 and ''s''2, we turn that slice into the bright generator, hence swapping ''s''''t''[-''g''] and ''s''''t''[-''g'' + 1]. Note that ''g'' must be odd since it generates a 2-period MOS. So (under 0-indexing) the first letter's index is even and the second letter's index is odd, which is what we want since the letters are within a stacked 2-step. While the generator might have to be higher by an (''n''/2)-step, that doesn't affect the parity since ''n''/2 is even.

We prove that ''s''1 and ''s''2 are MOS substitution scales with a filling MOS of period 2. The number the 2-step (1) occurs must be the same in both ''s''1 and ''s''2. The word of stacked 2-steps of the template MOS (which is of the form {{nowrap|''w''('''x''', '''X''', '''X''')''w''('''x''', '''X''', '''X''')}}), which is itself a MOS word, consists of letters (1) '''x''' + '''X''' and (2) 2'''X''' if more '''X''''s than '''x''''s, 2'''x''' if more '''x''''s than '''X''''s. The word of stacked 2-steps from our chosen offset is also this same MOS word. Thus it remains to handle the cases (1) and (2) above. ~~IWhenever~~ the letter '''x''' + '''X''' is encountered, the number of the last letters that are equated to '''X''' that are consumed is 1, which is odd. Whenever the other letter is encountered, that number is even (0 or 2). Hence (since ''n'' > 4) the letter 2'''X''' resp. 2'''x''' serves as the non-slot letter, and the letters ('''x''' + '''X''') serve as the slot letters where a 2-period filling MOS word (a repetition of {{nowrap|('''x'''+'''y''')('''x'''+'''z''')}}) is substituted.

We prove that ''s''1 and ''s''2 are MOS substitution scales with a filling MOS of period 2. The number the 2-step (1) occurs must be the same in both ''s''1 and ''s''2. The word of stacked 2-steps of the template MOS (which is of the form {{nowrap|''w''('''x''', '''X''', '''X''')''w''('''x''', '''X''', '''X''')}}), which is itself a MOS word, consists of letters (1) '''x''' + '''X''' and (2) 2'''X''' if more '''X''''s than '''x''''s, 2'''x''' if more '''x''''s than '''X''''s. The word of stacked 2-steps from our chosen offset is also this same MOS word. Thus it remains to handle the cases (1) and (2) above. Whenever the letter '''x''' + '''X''' is encountered, the number of the last letters that are equated to '''X''' that are consumed is 1, which is odd. Whenever the other letter is encountered, that number is even (0 or 2). Hence (since ''n'' > 4) the letter 2'''X''' resp. 2'''x''' serves as the non-slot letter, and the letters ('''x''' + '''X''') serve as the slot letters where a 2-period filling MOS word (a repetition of {{nowrap|('''x'''+'''y''')('''x'''+'''z''')}}) is substituted.

Now we count the letters that occur in these MOS substitution words of 2-steps. Consider the chunk boundaries of the template MOS. For every boundary between chunks, there is one slot letter in the template MOS for ''s''1 and one in the template MOS ''s''2, due to index parity. So it suffices that we have evenly many boundaries between (nonempty) chunks. Equivalently, we have to prove that there are evenly many steps of the step size that occurs less frequently in the template MOS ''a'''''x''' 2''k'''''X''', which is true by assumption (''a'' and 2''k'' are both even).

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 * In the doubly even case, start from the mode of ''s'' whose template MOS is the brightest mode. Let ''s''<sub>2</sub> be offset at a generator of the even-regular scale, which by Theorem 6 we choose to have the same interval class as a bright generator of the MOS ''a'''''x''' 2''k'''''X'''. This is what induces the equality of ''s''<sub>1</sub> and ''s''<sub>2</sub> (in particular, the two scales have the same period, thus they are both primitive): Let ''s''<sub>''t''</sub> be the period of the brightest mode of the template MOS, and let ''g'' be its bright generator class. Then the slice {{nowrap|''s''<sub>''t''</sub>[-''g'' +1 : 1]}} is the imperfect generator of the MOS. Now when we "darken" the mode by one generator, which is the difference between the template MOSes of ''s''<sub>1</sub> and ''s''<sub>2</sub>, we turn that slice into the bright generator, hence swapping ''s''<sub>''t''</sub>[-''g''] and ''s''<sub>''t''</sub>[-''g'' + 1]. Note that ''g'' must be odd since it generates a 2-period MOS. So (under 0-indexing) the first letter's index is even and the second letter's index is odd, which is what we want since the letters are within a stacked 2-step. While the generator might have to be higher by an (''n''/2)-step, that doesn't affect the parity since ''n''/2 is even.
-We prove that ''s''<sub>1</sub> and ''s''<sub>2</sub> are MOS substitution scales with a filling MOS of period 2. The number the 2-step (1) occurs must be the same in both ''s''<sub>1</sub> and ''s''<sub>2</sub>. The word of stacked 2-steps of the template MOS (which is of the form {{nowrap|''w''('''x''', '''X''', '''X''')''w''('''x''', '''X''', '''X''')}}), which is itself a MOS word, consists of letters (1) '''x''' + '''X''' and (2) 2'''X''' if more '''X''''s than '''x''''s, 2'''x''' if more '''x''''s than '''X''''s. The word of stacked 2-steps from our chosen offset is also this same MOS word. Thus it remains to handle the cases (1) and (2) above. IWhenever the letter '''x''' + '''X''' is encountered, the number of the last letters that are equated to '''X''' that are consumed is 1, which is odd. Whenever the other letter is encountered, that number is even (0 or 2). Hence (since ''n'' > 4) the letter 2'''X''' resp. 2'''x''' serves as the non-slot letter, and the letters ('''x''' + '''X''') serve as the slot letters where a 2-period filling MOS word (a repetition of {{nowrap|('''x'''+'''y''')('''x'''+'''z''')}}) is substituted.
+We prove that ''s''<sub>1</sub> and ''s''<sub>2</sub> are MOS substitution scales with a filling MOS of period 2. The number the 2-step (1) occurs must be the same in both ''s''<sub>1</sub> and ''s''<sub>2</sub>. The word of stacked 2-steps of the template MOS (which is of the form {{nowrap|''w''('''x''', '''X''', '''X''')''w''('''x''', '''X''', '''X''')}}), which is itself a MOS word, consists of letters (1) '''x''' + '''X''' and (2) 2'''X''' if more '''X''''s than '''x''''s, 2'''x''' if more '''x''''s than '''X''''s. The word of stacked 2-steps from our chosen offset is also this same MOS word. Thus it remains to handle the cases (1) and (2) above. Whenever the letter '''x''' + '''X''' is encountered, the number of the last letters that are equated to '''X''' that are consumed is 1, which is odd. Whenever the other letter is encountered, that number is even (0 or 2). Hence (since ''n'' > 4) the letter 2'''X''' resp. 2'''x''' serves as the non-slot letter, and the letters ('''x''' + '''X''') serve as the slot letters where a 2-period filling MOS word (a repetition of {{nowrap|('''x'''+'''y''')('''x'''+'''z''')}}) is substituted.
 Now we count the letters that occur in these MOS substitution words of 2-steps. Consider the chunk boundaries of the template MOS. For every boundary between chunks, there is one slot letter in the template MOS for ''s''<sub>1</sub> and one in the template MOS ''s''<sub>2</sub>, due to index parity. So it suffices that we have evenly many boundaries between (nonempty) chunks. Equivalently, we have to prove that there are evenly many steps of the step size that occurs less frequently in the template MOS ''a'''''x''' 2''k'''''X''', which is true by assumption (''a'' and 2''k'' are both even).