Ternary scale theorems: Difference between revisions

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These two claims prove that {{nowrap|''E'''''X'''(S) {{=}} ('''YZ''')''b''}} and that the two GS generators' sizes differ by replacing one '''Y''' for a '''Z'''. {{Qed}}

== Theorem 2 ~~(Odd generator-offset scales are AGS) ==~~

== Theorem 2 (Classification of pairwise well-formed scales) ==

~~Suppose that a periodic scale satisfies the following:~~

* is generator-offset

* has odd size ''n''.

~~Then the scale is AGS.~~

~~=== Proof ===~~

Assume that the generator '''g''' is a ''k''-step and ''k'' is even. (If ''k'' is not even, invert the generator.) On some note ''p'' we have a chain of (''n'' + 1)/2 notes and on ''p′'' {{=}} ''p'' + offset we'll have (''n'' − 1)/2) notes.

Assume 1 < gcd(''k'', ''n'') < ''n'' and ''n'' ≥ 5. Since ''n'' is odd, ''d'' {{=}} gcd(''k'', ''n'') is an odd number at least 3, and by well-formedness with respect to the generator, there must be a circle of ''n''/''d'' < {{floor|''n''/2}} notes formed by '''g''', contrary to the assumption of GO. Thus, gcd(''k'', ''n'') {{=}} 1.

Since ''n'' is odd, ''rk'' ≡ ''k''/2 mod ''n'' iff ''r'' ≡ (''n'' + 1)/2 mod ''n''. (Note that both 2 and ''k'' are coprime with ''n'', hence multiplicatively invertible mod ''n''.) This proves that the offset, which must be reached after (''n'' + 1)/2 ''k''-steps, is a ''k''/2-step, as desired. (As [''k''] is a generator of ℤ/''n'', stacking (''n'' − 1)-many ''k''-steps must visit every note exactly once. Thus if the offset wasn't reached in (''n'' + 1)/2 steps, the two generator chains wouldn't have the assumed lengths.) {{qed}}

~~== Theorem 3 (Properties of even generator-offset ternary scales) ==~~

~~A primitive generator-offset ternary scale ''s'' of even size 6 or greater, where the generator '''g''' is an even-step, has the following properties:~~

~~# ''s'' is a union of two copies of a primitive MOS ''M'' of size {{sfrac|''n''|2}} generated by '''g'''; thus it is an [[interleaving]] obtained by taking two offset copies of said primitive MOS.~~

~~# ''s'' is ''not'' SV3.~~

~~# ''s'' is ''not'' chiral.~~

# If {{nowrap|''M'' {{=}} ''M''('''y''', '''z''')}} is the primitive MOS necklace above, then {{nowrap|''s'' {{=}} ''M''('''XY''', '''XZ''')}} for some assignment of variable names '''X''', '''Y''', and '''Z''' to the three letters of ''s''.

~~=== Proof ===~~

(1) and (2) were proved in the proof of Proposition 1 (the part that we appeal to, from "all multiples of the generator '''g''' must be even-steps ..." to "These are all distinct by ℤ-linear independence", does not rely on ''s'' having the AGS property). (3) and (4) are easy to check using (1). {{qed}}

~~== Theorem 4~~ (Classification of pairwise well-formed scales) ==

Let {{nowrap|''s''('''a''', '''b''', '''c''')}} be a scale word in three ℤ-linearly independent step sizes '''a''', '''b''', '''c'''. Suppose ''s'' is pairwise well-formed (equivalently, all its projections are primitive MOSes). Then ''s'' is SV3 and has an odd number of notes. Moreover, ''s'' is either generator-offset or equivalent to the scale word '''abacaba'''.

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For any individual letter '''X''', identify letters other than it to get a MOS. Since MOS words are balanced, the block balance for any letter '''X''' is at most 1, as required by the balance property. {{Qed}}

== Theorem 6 (Generator-offset structure of even-regular scales) ==

== Theorem 3 (Generator-offset structure of even-regular scales) ==

=== Definition (Even-regular scale) ===

A primitive ternary scale ''s'' is ''even-regular'' if len(''s'') is even and ''s'' is equivalent to a word constructed from taking the MOS 2''a'''''X''' 2''c'''''Z''' with ''a'' odd and {{nowrap|gcd(''a'', ''c'') {{=}} 1}}, and replacing every other '''X''' with '''Y'''. In particular, ''s'' has [[step signature]] equivalent to ''a'''''X''' ''a'''''Y''' ''b'''''Z''' with ''a'' odd and ''b'' even. For example, '''LsLsLmsLsLsm''' (achiral [[diachrome]], 5'''L''' 2'''m''' 5'''s''') is an even-regular scale.

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It remains to show that ''s'' is balanced. Any ''k''-step subword has either ''j'' or ''j'' + 1 '''Z'''s for some ''j'' since the result of conflating '''X''' and '''Y''' is a MOS, and ''k''-step subwords for both possibilities exist when 0 < ''k'' < len(''s'')/2. If the number of non-'''Z''' letters in a ''k''-step subword is even, then there is only one possibility for the number of '''X''' and the number of '''Y'''. If the number of non-'''Z''' letters in a ''k''-step subword is odd, then both the number of '''X'''s and the number of '''Y'''s differ by at most 1. {{qed}}

== Theorem 7 (Classification of MV3 scales) ==

== Theorem 4 (Classification of MV3 scales) ==

In the following, ''equivalent'' means "is the same circular word after permuting '''X''', '''Y''', and '''Z'''." This means that '''XYXZXYX''' is equivalent to '''YZYXYZY''', or '''XZXYXZX''', and so on.

=== Theorem 7.1 (Classification of ternary balanced scales) ===

=== Theorem 4.1 (Classification of ternary balanced scales) ===

# A primitive [[balanced]] ternary scale ''s'' is pairwise-MOS, and satisfies one of the following:

## '''sporadic balanced''': ''s'' is equivalent to '''XYXZXYX''', the ternary [[Fraenkel word]], with step signature 4'''X'''2'''Y'''1'''Z'''.

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==== Proof ====

For 7.1.1: We showed previously that the Fraenkel, odd-regular, and even-regular circular words are balanced. Thus it remains to show that (a) ternary balanced words are pairwise-MOS (b) if ''a'' > ''b'' > ''c'', then ''s'' is equivalent to the Fraenkel word (c) assuming ''a'' != ''b'' = ''c'' any ''s'' that is not odd-regular or even-regular is not balanced.

For 4.1.1: We showed previously that the Fraenkel, odd-regular, and even-regular circular words are balanced. Thus it remains to show that (a) ternary balanced words are pairwise-MOS (b) if ''a'' > ''b'' > ''c'', then ''s'' is equivalent to the Fraenkel word (c) assuming ''a'' != ''b'' = ''c'' any ''s'' that is not odd-regular or even-regular is not balanced.

(a) Let ''s'' be a ternary balanced word; then for any given letter '''y''' the number of '''y'''s in a subword of any given length ''L'' varies by at most 1. Thus the same is true when we count all non-'''y''' letters in any subword of length ''L''; thus when we equate '''x''' and '''z''', the count of the resulting letter in any subword of length ''L'' differs by 1. Being a binary balanced word is one characterization of the MOS property.

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(c) The scale made by taking ''s'' and conflating '''Y''' and '''Z''' into the letter '''W''' must be a MOS. To this scale we may imagine substituting a scale made of an equal amount of '''Y''' and '''Z''' letters into the "slot letters" '''W''' letter by letter. Let ''t''1 be a length-''k'' subword of the form '''YX'''''k''-2'''Y''' under the projection. We may assume that the chunk sizes of the MOS are ''k'' - 2 and ''k'' - 1, or ''k'' - 2 and ''k'' - 3. Either way, there exists some subword with (''k'' - i)-many '''X'''s, i = 1 or 2, and two '''Z'''s. This violates balance because ''t''1 contains zero '''Z'''s.

For 7.1.2: Suppose ''s'' is balanced and has at least three sizes for ''k''-steps, {{nowrap|''a''''i'''''X''' + ''b''''i'''''Y''' + ''c''''i'''''Z''' {{=}} (''a''''i'', ''b''''i'', ''c''''i'')}} for {{nowrap|''i'' ∈ {{(}}1, 2, 3{{)}}}}. We may assume {{nowrap|(''a''2, ''b''2, ''c''2) {{=}} (''a''1, ''b''1 + 1, ''c''1 − 1)}}. Then either {{nowrap|(''a''3, ''b''3, ''c''3) {{=}} (''a''1 + 1, ''b''1, ''c''1 − 1)}} or {{nowrap|(''a''3, ''b''3, ''c''3) {{=}} (''a''1 − 1, ''b''1 + 1, ''c''1)}}. In both cases, by balancedness applied to subwords of length ''k'', the three vectors represent the only possible interval sizes.

For 4.1.2: Suppose ''s'' is balanced and has at least three sizes for ''k''-steps, {{nowrap|''a''''i'''''X''' + ''b''''i'''''Y''' + ''c''''i'''''Z''' {{=}} (''a''''i'', ''b''''i'', ''c''''i'')}} for {{nowrap|''i'' ∈ {{(}}1, 2, 3{{)}}}}. We may assume {{nowrap|(''a''2, ''b''2, ''c''2) {{=}} (''a''1, ''b''1 + 1, ''c''1 − 1)}}. Then either {{nowrap|(''a''3, ''b''3, ''c''3) {{=}} (''a''1 + 1, ''b''1, ''c''1 − 1)}} or {{nowrap|(''a''3, ''b''3, ''c''3) {{=}} (''a''1 − 1, ''b''1 + 1, ''c''1)}}. In both cases, by balancedness applied to subwords of length ''k'', the three vectors represent the only possible interval sizes.

For 7.1.3: The ternary Fraenkel word may be verified as SV3 by inspection, and we have already shown in Theorem 1 that odd-regular balanced scales are SV3. To show that even-regular balanced scales are ''not'' SV3, observe that {{nowrap|(''a'' + ''c'')}}-steps come in only 2 sizes in such a scale ''s'': {{nowrap|{{floor|''a''/2}}'''X''' + {{ceil|''a''/2}}'''Y''' + ''c'''''Z'''}} and {{nowrap|{{ceil|''a''/2}}'''X''' + {{floor|''a''/2}}'''Y''' + ''c'''''Z'''}}, since the underlying MOS 2''a'''''X'''2''c'''''Y''' only has the {{nowrap|(''a'' + ''c'')}}-step {{nowrap|''a'''''X''' + ''c'''''Z'''}}. The construction replaces the '''X'''s in these subwords with alternating '''X'''s and '''Y'''s; either of '''X''' or '''Y''' may occur first, corresponding to the two possible sizes, since ''a'' is odd and thus the {{nowrap|(''a'' + ''c'')}}-step subword {{nowrap|''s''[''k'' − 1 : ''k'' + ''a'' + ''c'' − 1]}} becomes the subword {{nowrap|''s''[''k'' + ''a'' + ''c'' − 1 : ''k'' + 2''a'' + 2''c'' − 1]}} via interchanging '''X''' and '''Y'''.

For 4.1.3: The ternary Fraenkel word may be verified as SV3 by inspection, and we have already shown in Theorem 1 that odd-regular balanced scales are SV3. To show that even-regular balanced scales are ''not'' SV3, observe that {{nowrap|(''a'' + ''c'')}}-steps come in only 2 sizes in such a scale ''s'': {{nowrap|{{floor|''a''/2}}'''X''' + {{ceil|''a''/2}}'''Y''' + ''c'''''Z'''}} and {{nowrap|{{ceil|''a''/2}}'''X''' + {{floor|''a''/2}}'''Y''' + ''c'''''Z'''}}, since the underlying MOS 2''a'''''X'''2''c'''''Y''' only has the {{nowrap|(''a'' + ''c'')}}-step {{nowrap|''a'''''X''' + ''c'''''Z'''}}. The construction replaces the '''X'''s in these subwords with alternating '''X'''s and '''Y'''s; either of '''X''' or '''Y''' may occur first, corresponding to the two possible sizes, since ''a'' is odd and thus the {{nowrap|(''a'' + ''c'')}}-step subword {{nowrap|''s''[''k'' − 1 : ''k'' + ''a'' + ''c'' − 1]}} becomes the subword {{nowrap|''s''[''k'' + ''a'' + ''c'' − 1 : ''k'' + 2''a'' + 2''c'' − 1]}} via interchanging '''X''' and '''Y'''.

Claim 7.1.4 can be verified by noting that such scales are PWF and using Theorem 4. {{Qed}}

Claim 4.1.4 can be verified by noting that such scales are PWF and using Theorem 4. {{Qed}}

=== Theorem 7.2 (Classification of MV3 scales) ===

=== Theorem 4.2 (Classification of MV3 scales) ===

A primitive MV3 scale is either

# '''balanced''' (classified by the previous theorem),

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Note: The xen term "brightest MOS word" is equivalent to "Christoffel word" in the paper, and similarly "brightest multiMOS word" is equivalent to "powers of a Christoffel word". Also see [[Glossary for combinatorics on words]] for more equivalents between xen community terms and standard academic terminology.

== Theorem 8 (Even-regular scales as (contra)interleavings) ==

== Theorem 5 (Even-regular scales as (contra)interleavings) ==

Let ''s'' be a primitive even-regular scale of [[MOS substitution]] type ''a'''''x'''(''k'''''y''' ''k'''''z''') where ''a'' is even and gcd(''a'', ''k'') = 1. Let ''n'' = |''s''| = ''a'' + 2''k''.

# If ''n'' is singly even, then ''s'' is a [[interleaving|contrainterleaving]] of the two opposite chiralities of an odd-regular scale.

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 These two claims prove that {{nowrap|''E''<sub>'''X'''</sub>(S) {{=}} ('''YZ''')<sup>''b''</sup>}} and that the two GS generators' sizes differ by replacing one '''Y''' for a '''Z'''. {{Qed}}
-== Theorem 2 (Odd generator-offset scales are AGS) ==
+== Theorem 2 (Classification of pairwise well-formed scales) ==
-Suppose that a periodic scale satisfies the following:
-* is generator-offset
-* has odd size ''n''.
-Then the scale is AGS.
-=== Proof ===
-Assume that the generator '''g''' is a ''k''-step and ''k'' is even. (If ''k'' is not even, invert the generator.) On some note ''p'' we have a chain of (''n'' + 1)/2 notes and on ''p′'' {{=}} ''p'' + offset we'll have (''n'' − 1)/2) notes.
-Assume 1 &lt; gcd(''k'', ''n'') &lt; ''n'' and ''n'' &ge; 5. Since ''n'' is odd, ''d'' {{=}} gcd(''k'', ''n'') is an odd number at least 3, and by well-formedness with respect to the generator, there must be a circle of ''n''/''d'' &lt; {{floor|''n''/2}} notes formed by '''g''', contrary to the assumption of GO. Thus, gcd(''k'', ''n'') {{=}} 1.
-Since ''n'' is odd, ''rk'' ≡ ''k''/2 mod ''n'' iff ''r'' ≡ (''n'' + 1)/2 mod ''n''. (Note that both 2 and ''k'' are coprime with ''n'', hence multiplicatively invertible mod ''n''.) This proves that the offset, which must be reached after (''n'' + 1)/2 ''k''-steps, is a ''k''/2-step, as desired. (As [''k''] is a generator of ℤ/''n'', stacking (''n'' − 1)-many ''k''-steps must visit every note exactly once. Thus if the offset wasn't reached in (''n'' + 1)/2 steps, the two generator chains wouldn't have the assumed lengths.) {{qed}}
-== Theorem 3 (Properties of even generator-offset ternary scales) ==
-A primitive generator-offset ternary scale ''s'' of even size 6 or greater, where the generator '''g''' is an even-step, has the following properties:
-# ''s'' is a union of two copies of a primitive MOS ''M'' of size {{sfrac|''n''|2}} generated by '''g'''; thus it is an [[interleaving]] obtained by taking two offset copies of said primitive MOS.
-# ''s'' is ''not'' SV3.
-# ''s'' is ''not'' chiral.
-# If {{nowrap|''M'' {{=}} ''M''('''y''', '''z''')}} is the primitive MOS necklace above, then {{nowrap|''s'' {{=}} ''M''('''XY''', '''XZ''')}} for some assignment of variable names '''X''', '''Y''', and '''Z''' to the three letters of ''s''.
-=== Proof ===
-(1) and (2) were proved in the proof of Proposition 1 (the part that we appeal to, from "all multiples of the generator '''g''' must be even-steps ..." to "These are all distinct by ℤ-linear independence", does not rely on ''s'' having the AGS property). (3) and (4) are easy to check using (1). {{qed}}
-== Theorem 4 (Classification of pairwise well-formed scales) ==
 Let {{nowrap|''s''('''a''', '''b''', '''c''')}} be a scale word in three ℤ-linearly independent step sizes '''a''', '''b''', '''c'''. Suppose ''s'' is pairwise well-formed (equivalently, all its projections are primitive MOSes). Then ''s'' is SV3 and has an odd number of notes. Moreover, ''s'' is either generator-offset or equivalent to the scale word '''abacaba'''.
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 For any individual letter '''X''', identify letters other than it to get a MOS. Since MOS words are balanced, the block balance for any letter '''X''' is at most 1, as required by the balance property. {{Qed}}
-== Theorem 6 (Generator-offset structure of even-regular scales) ==
+== Theorem 3 (Generator-offset structure of even-regular scales) ==
 === Definition (Even-regular scale) ===
 A primitive ternary scale ''s'' is ''even-regular'' if len(''s'') is even and ''s'' is equivalent to a word constructed from taking the MOS 2''a'''''X'''&nbsp;2''c'''''Z''' with ''a'' odd and {{nowrap|gcd(''a'', ''c'') {{=}} 1}}, and replacing every other '''X''' with '''Y'''. In particular,  ''s'' has [[step signature]] equivalent to ''a'''''X'''&nbsp;''a'''''Y'''&nbsp;''b'''''Z''' with ''a'' odd and ''b'' even. For example, '''LsLsLmsLsLsm''' (achiral [[diachrome]], 5'''L'''&nbsp;2'''m'''&nbsp;5'''s''') is an even-regular scale.
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 It remains to show that ''s'' is balanced. Any ''k''-step subword has either ''j'' or ''j'' + 1 '''Z'''s for some ''j'' since the result of conflating '''X''' and '''Y''' is a MOS, and ''k''-step subwords for both possibilities exist when 0 < ''k'' < len(''s'')/2. If the number of non-'''Z''' letters in a ''k''-step subword is even, then there is only one possibility for the number of '''X''' and the number of '''Y'''. If the number of non-'''Z''' letters in a ''k''-step subword is odd, then both the number of '''X'''s and the number of '''Y'''s differ by at most 1. {{qed}}
-== Theorem 7 (Classification of MV3 scales) ==
+== Theorem 4 (Classification of MV3 scales) ==
 In the following, ''equivalent'' means "is the same circular word after permuting '''X''', '''Y''', and '''Z'''." This means that '''XYXZXYX''' is equivalent to '''YZYXYZY''', or '''XZXYXZX''', and so on.
-=== Theorem 7.1 (Classification of ternary balanced scales) ===
+=== Theorem 4.1 (Classification of ternary balanced scales) ===
 # A primitive [[balanced]] ternary scale ''s'' is pairwise-MOS, and satisfies one of the following:
 ## '''sporadic balanced''': ''s'' is equivalent to '''XYXZXYX''', the ternary [[Fraenkel word]], with step signature 4'''X'''2'''Y'''1'''Z'''.
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 ==== Proof ====
-For 7.1.1: We showed previously that the Fraenkel, odd-regular, and even-regular circular words are balanced. Thus it remains to show that (a) ternary balanced words are pairwise-MOS (b) if ''a'' > ''b'' > ''c'', then ''s'' is equivalent to the Fraenkel word (c) assuming ''a'' != ''b'' = ''c'' any ''s'' that is not odd-regular or even-regular is not balanced.
+For 4.1.1: We showed previously that the Fraenkel, odd-regular, and even-regular circular words are balanced. Thus it remains to show that (a) ternary balanced words are pairwise-MOS (b) if ''a'' > ''b'' > ''c'', then ''s'' is equivalent to the Fraenkel word (c) assuming ''a'' != ''b'' = ''c'' any ''s'' that is not odd-regular or even-regular is not balanced.
 (a) Let ''s'' be a ternary balanced word; then for any given letter '''y''' the number of '''y'''s in a subword of any given length ''L'' varies by at most 1. Thus the same is true when we count all non-'''y''' letters in any subword of length ''L''; thus when we equate '''x''' and '''z''', the count of the resulting letter in any subword of length ''L'' differs by 1. Being a binary balanced word is one characterization of the MOS property.
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 (c) The scale made by taking ''s'' and conflating '''Y''' and '''Z''' into the letter '''W''' must be a MOS. To this scale we may imagine substituting a scale made of an equal amount of '''Y''' and '''Z''' letters into the "slot letters" '''W''' letter by letter. Let ''t''<sub>1</sub> be a length-''k'' subword of the form '''YX'''<sup>''k''-2</sup>'''Y''' under the projection. We may assume that the chunk sizes of the MOS are ''k'' - 2 and ''k'' - 1, or ''k'' - 2 and ''k'' - 3. Either way, there exists some subword with (''k'' - i)-many '''X'''s, i = 1 or 2, and two '''Z'''s. This violates balance because ''t''<sub>1</sub> contains zero '''Z'''s.
-For 7.1.2: Suppose ''s'' is balanced and has at least three sizes for ''k''-steps, {{nowrap|''a''<sub>''i''</sub>'''X''' + ''b''<sub>''i''</sub>'''Y''' + ''c''<sub>''i''</sub>'''Z''' {{=}} (''a''<sub>''i''</sub>, ''b''<sub>''i''</sub>, ''c''<sub>''i''</sub>)}} for {{nowrap|''i'' ∈ {{(}}1, 2, 3{{)}}}}. We may assume {{nowrap|(''a''<sub>2</sub>, ''b''<sub>2</sub>, ''c''<sub>2</sub>) {{=}} (''a''<sub>1</sub>, ''b''<sub>1</sub> + 1, ''c''<sub>1</sub> − 1)}}. Then either {{nowrap|(''a''<sub>3</sub>, ''b''<sub>3</sub>, ''c''<sub>3</sub>) {{=}} (''a''<sub>1</sub> + 1, ''b''<sub>1</sub>, ''c''<sub>1</sub> − 1)}} or {{nowrap|(''a''<sub>3</sub>, ''b''<sub>3</sub>, ''c''<sub>3</sub>) {{=}} (''a''<sub>1</sub> − 1, ''b''<sub>1</sub> + 1, ''c''<sub>1</sub>)}}. In both cases, by balancedness applied to subwords of length ''k'', the three vectors represent the only possible interval sizes.
+For 4.1.2: Suppose ''s'' is balanced and has at least three sizes for ''k''-steps, {{nowrap|''a''<sub>''i''</sub>'''X''' + ''b''<sub>''i''</sub>'''Y''' + ''c''<sub>''i''</sub>'''Z''' {{=}} (''a''<sub>''i''</sub>, ''b''<sub>''i''</sub>, ''c''<sub>''i''</sub>)}} for {{nowrap|''i'' ∈ {{(}}1, 2, 3{{)}}}}. We may assume {{nowrap|(''a''<sub>2</sub>, ''b''<sub>2</sub>, ''c''<sub>2</sub>) {{=}} (''a''<sub>1</sub>, ''b''<sub>1</sub> + 1, ''c''<sub>1</sub> − 1)}}. Then either {{nowrap|(''a''<sub>3</sub>, ''b''<sub>3</sub>, ''c''<sub>3</sub>) {{=}} (''a''<sub>1</sub> + 1, ''b''<sub>1</sub>, ''c''<sub>1</sub> − 1)}} or {{nowrap|(''a''<sub>3</sub>, ''b''<sub>3</sub>, ''c''<sub>3</sub>) {{=}} (''a''<sub>1</sub> − 1, ''b''<sub>1</sub> + 1, ''c''<sub>1</sub>)}}. In both cases, by balancedness applied to subwords of length ''k'', the three vectors represent the only possible interval sizes.
-For 7.1.3: The ternary Fraenkel word may be verified as SV3 by inspection, and we have already shown in Theorem 1 that odd-regular balanced scales are SV3. To show that even-regular balanced scales are ''not'' SV3, observe that {{nowrap|(''a'' + ''c'')}}-steps come in only 2 sizes in such a scale ''s'': {{nowrap|{{floor|''a''/2}}'''X''' + {{ceil|''a''/2}}'''Y''' + ''c'''''Z'''}} and {{nowrap|{{ceil|''a''/2}}'''X''' + {{floor|''a''/2}}'''Y''' + ''c'''''Z'''}}, since the underlying MOS 2''a'''''X'''2''c'''''Y''' only has the {{nowrap|(''a'' + ''c'')}}-step {{nowrap|''a'''''X''' + ''c'''''Z'''}}. The construction replaces the '''X'''s in these subwords with alternating '''X'''s and '''Y'''s; either of '''X''' or '''Y''' may occur first, corresponding to the two possible sizes, since ''a'' is odd and thus the {{nowrap|(''a'' + ''c'')}}-step subword {{nowrap|''s''[''k'' &minus; 1 : ''k'' + ''a'' + ''c'' &minus; 1]}} becomes the subword {{nowrap|''s''[''k'' + ''a'' + ''c'' &minus; 1 : ''k'' + 2''a'' + 2''c'' &minus; 1]}} via interchanging '''X''' and '''Y'''.
+For 4.1.3: The ternary Fraenkel word may be verified as SV3 by inspection, and we have already shown in Theorem 1 that odd-regular balanced scales are SV3. To show that even-regular balanced scales are ''not'' SV3, observe that {{nowrap|(''a'' + ''c'')}}-steps come in only 2 sizes in such a scale ''s'': {{nowrap|{{floor|''a''/2}}'''X''' + {{ceil|''a''/2}}'''Y''' + ''c'''''Z'''}} and {{nowrap|{{ceil|''a''/2}}'''X''' + {{floor|''a''/2}}'''Y''' + ''c'''''Z'''}}, since the underlying MOS 2''a'''''X'''2''c'''''Y''' only has the {{nowrap|(''a'' + ''c'')}}-step {{nowrap|''a'''''X''' + ''c'''''Z'''}}. The construction replaces the '''X'''s in these subwords with alternating '''X'''s and '''Y'''s; either of '''X''' or '''Y''' may occur first, corresponding to the two possible sizes, since ''a'' is odd and thus the {{nowrap|(''a'' + ''c'')}}-step subword {{nowrap|''s''[''k'' &minus; 1 : ''k'' + ''a'' + ''c'' &minus; 1]}} becomes the subword {{nowrap|''s''[''k'' + ''a'' + ''c'' &minus; 1 : ''k'' + 2''a'' + 2''c'' &minus; 1]}} via interchanging '''X''' and '''Y'''.
-Claim 7.1.4 can be verified by noting that such scales are PWF and using Theorem 4. {{Qed}}
+Claim 4.1.4 can be verified by noting that such scales are PWF and using Theorem 4. {{Qed}}
-=== Theorem 7.2 (Classification of MV3 scales) ===
+=== Theorem 4.2 (Classification of MV3 scales) ===
 A primitive MV3 scale is either
 # '''balanced''' (classified by the previous theorem),
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 Note: The xen term "brightest MOS word" is equivalent to "Christoffel word" in the paper, and similarly "brightest multiMOS word" is equivalent to "powers of a Christoffel word". Also see [[Glossary for combinatorics on words]] for more equivalents between xen community terms and standard academic terminology.
-== Theorem 8 (Even-regular scales as (contra)interleavings) ==
+== Theorem 5 (Even-regular scales as (contra)interleavings) ==
 Let ''s'' be a primitive even-regular scale of [[MOS substitution]] type ''a'''''x'''(''k'''''y''' ''k'''''z''') where ''a'' is even and gcd(''a'', ''k'') = 1. Let ''n'' = |''s''| = ''a'' + 2''k''.
 # If ''n'' is singly even, then ''s'' is a [[interleaving|contrainterleaving]] of the two opposite chiralities of an odd-regular scale.