Rearrange the digits in ⟨125034⟩ to meet the rules below.
⟨5th 4th 3rd 2nd 1st 0th⟩
✅Match
⟨⋯ 3 ⋯ ? ⋯ 1 (?−2)⟩ (?≠3)
⛔Avoid
max {p5, p1, p0} = 5
5th → 0|1|2|3
⟨⋯ ? ⋯ 2 ⋯ (?−5)⟩
#125034_v2.10
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│▒
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Step 1 │ │ │ │ │ 1 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ 4 │ │ │ │ 1 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 4 │ │ │ │ 1 │ 0 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 4 │ 3 │ │ │ 1 │ 0 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 4 │ 3 │ 2 │ │ 1 │ 0 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 4 │ 3 │ 2 │ 5 │ 1 │ 0 │▒
└───┴───┴───┴───┴───┴───┘▒
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Proof of 2025-10-21 WR
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Notation: if nth -> a, then we write [nth] = a.
Plainly, by ✅「⟨⋯ 3 ⋯ ? ⋯ 1 (?−2)⟩ (?≠3)」, we get [1st] = 1:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│ 1■│0th│▒
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Step 1 │ │ │ │ │ 1 │ │▒
└───┴───┴───┴───┴───┴───┘▒
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--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ 5 │ 0 │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
On the other hand, to avoid ⛔「5th → 0|1|2|3」, we need [5th] = 4 or 5. If it is 5 then we will match ⛔「max {p5, p1, p0} = 5」, which is a contradiction. Therefore, [5th] = 4:
┌───┬───┬───┬───┬───┬───┐
│ 5■│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ │ │ 1 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ 4 │ │ │ │ 1 │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ 5 │ 0 │ 3 │ │
└───┴───┴───┴───┴───┴───┘
Next, we use ✅「⟨⋯ 3 ⋯ ? ⋯ 1 (?−2)⟩ (?≠3)」 to deduce what [0th] is. This pattern implies that [0th] <= 5-2 = 3, or [0th] = 0|2|3. The pattern requires that 3 is to the left of 1, so [0th] is not 3. If it is 2, then to match the pattern, we need 3 is to the left of 4, which is not possible. Therefore, [0th] = 0:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 4 │ │ │ │ 1 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 4 │ │ │ │ 1 │ 0 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ 5 │ │ 3 │ │
└───┴───┴───┴───┴───┴───┘
As a result, the "?" in ✅「⟨⋯ 3 ⋯ ? ⋯ 1 (?−2)⟩ (?≠3)」 is 2, and 3 is to the left of 2. On the other hand, to avoid ⛔「⟨⋯ ? ⋯ 2 ⋯ (?−5)⟩」, we need 2 is to the left of 5. It follows that we have to match the pattern ⟨⋯ 3 ⋯ 2 ⋯ 5 ⋯⟩. We finish by
┌───┬───┬───┬───┬───┬───┐
│5th│ 4■│ 3■│ 2■│1st│0th│▒
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│ 4 │ │ │ │ 1 │ 0 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 4 │ 3 │ │ │ 1 │ 0 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 4 │ 3 │ 2 │ │ 1 │ 0 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 4 │ 3 │ 2 │ 5 │ 1 │ 0 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.10