Generator-offset property: Difference between revisions

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(Note: We say “assume” as shorthand for either “assume without loss of generality” or “assume after excluding other possibilities”.)

Suppose that S2's generator is also a k-step, and that Σ2's I is located at index n. Then Σ1 and Σ2 are the same mos pattern (up to knowing which step size is the bigger one) and even the same mode:

Suppose for sake of contradiction that P has only one preimage in S. Assume that S2's generator is also a k-step, and that Σ2's imperfect generator is located at index n. Then S1 and S2 are the same mos pattern (up to knowing which step size is the bigger one) and even the same mode. Assume the L of S1 (it could be s, but it doesn’t matter) is the result of identifying b and c, and all instances of s in S1 come from a. Then the steps of S2 corresponding to the L of S1 must be either all b’s or all a~c’s, thus these steps are all b’s in S (otherwise they would be identified with the a, against the assumption that S1 and S2 are the same mos pattern and mode). So S has only two step sizes (a and b), contradicting the assumption that S has exactly three step sizes.

~~S1: L ... L s~~

~~S2: L' ..~~. ~~L' s'~~

Assume the L of S1 (it could be s, but it doesn’t matter) is the result of identifying b and c, and all instances of s in S1 come from a. Then the steps of S2 corresponding to the L of S1 must be either all b’s or all a~c’s, thus these steps are all b’s in S (otherwise they would be identified with the a, against the assumption that S1 and S2 are the same mos pattern and mode). So S has only two step sizes (a and b), contradicting the assumption that S has exactly three step sizes~~. This shows that the maximum variety of S must be at least 3, and that P has at least two preimages in S~~.

We can write sizes of intervals in S as vectors (p, q, r) using the basis (a, b, c). Only two k-steps of S can project to P in S1, for if P has three preimages (α, β, γ), (α, β’, γ’), (α, β’’, γ’’) in S, then β, β’ and β’’ are three distinct values. Thus these would project to three different k-steps in S3, contradicting the mos property of S3.

@@ Line 125: / Line 125: @@
 (Note: We say “assume” as shorthand for either “assume without loss of generality” or “assume after excluding other possibilities”.)
-Suppose that S<sub>2</sub>'s generator is also a k-step, and that Σ<sub>2</sub>'s I is located at index n. Then Σ<sub>1</sub> and Σ<sub>2</sub> are the same mos pattern (up to knowing which step size is the bigger one) and even the same mode:
+Suppose for sake of contradiction that P has only one preimage in S. Assume that S<sub>2</sub>'s generator is also a k-step, and that Σ<sub>2</sub>'s imperfect generator is located at index n. Then S<sub>1</sub> and S<sub>2</sub> are the same mos pattern (up to knowing which step size is the bigger one) and even the same mode. Assume the L of S<sub>1</sub> (it could be s, but it doesn’t matter) is the result of identifying b and c, and all instances of s in S<sub>1</sub> come from a. Then the steps of S<sub>2</sub> corresponding to the L of S<sub>1</sub> must be either all b’s or all a~c’s, thus these steps are all b’s in S (otherwise they would be identified with the a, against the assumption that S<sub>1</sub> and S<sub>2</sub> are the same mos pattern and mode). So S has only two step sizes (a and b), contradicting the assumption that S has exactly three step sizes.
- S<sub>1</sub>:    L  ... L  s
- S<sub>2</sub>:    L' ... L' s'
-Assume the L of S<sub>1</sub> (it could be s, but it doesn’t matter) is the result of identifying b and c, and all instances of s in S<sub>1</sub> come from a. Then the steps of S<sub>2</sub> corresponding to the L of S<sub>1</sub> must be either all b’s or all a~c’s, thus these steps are all b’s in S (otherwise they would be identified with the a, against the assumption that S<sub>1</sub> and S<sub>2</sub> are the same mos pattern and mode). So S has only two step sizes (a and b), contradicting the assumption that S has exactly three step sizes. This shows that the maximum variety of S must be at least 3, and that P has at least two preimages in S.
 We can write sizes of intervals in S as vectors (p, q, r) using the basis (a, b, c). Only two k-steps of S can project to P in S<sub>1</sub>, for if P has three preimages (α, β, γ), (α, β’, γ’), (α, β’’, γ’’) in S, then β, β’ and β’’ are three distinct values. Thus these would project to three different k-steps in S<sub>3</sub>, contradicting the mos property of S<sub>3</sub>.