Rearrange the digits in ⟨125034⟩ to meet the rules below.
⟨ ⁵ᵗʰ▨ ⁴ᵗʰ▨ ³ʳᵈ▨ ²ⁿᵈ▨ ¹ˢᵗ▨ ⁰ᵗʰ▨ ⟩
✅Match
⟨⋯ 0 ⋯ 3 ⋯ 1 ⋯⟩
⛔Avoid
⟨ ³ʳᵈa ¹ˢᵗb ⁰ᵗʰc ⟩, (abc)₁₀ ≥ 314
⟨ ⁰ᵗʰ↑ ⟩ after ⟨→⟩
0th → 2|3|5
5th|3rd|0th → 0
#125034_v2.12
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ │ │ 4 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ 0 │ │ │ │ 4 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 5 │ 0 │ │ │ │ 4 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 5 │ 0 │ │ │ 1 │ 4 │▒
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Step 5 │ 5 │ 0 │ 2 │ │ 1 │ 4 │▒
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Step 6 │ 5 │ 0 │ 2 │ 3 │ 1 │ 4 │▒
└───┴───┴───┴───┴───┴───┘▒
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Proof of 2026-03-10 WR
══════════════════════
Notation: if nth -> a, then we write [nth] = a.
To avoid ⛔「0th → 2|3|5」, we need
(1) [0th] = 0|1|4.
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ │ │ │ ▬ │
└───┴───┴───┴───┴───┴───┘
If it is 0, then we cannot match ✅「⟨⋯ 0 ⋯ 3 ⋯ 1 ⋯⟩」. Else if it is 1, then [1st] != 0 by the preceding pattern, so we have [1st] > 1 = [0th]. Consequently, we would match ⛔「⟨ ⁰ᵗʰ↑ ⟩ after ⟨→⟩」, which is a contradiction. Hence, we see that [0th] = 4:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ │ │ 4 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
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│ 1 │ 2 │ 5 │ 0 │ 3 │ │
└───┴───┴───┴───┴───┴───┘
Next, we consider where to place 0. In view of ⛔「5th|3rd|0th → 0」, we have
(2) 0 = [4th] | [2nd] | [1st].
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│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ ▬ │ │ ▬ │ ▬ │ 4 │
└───┴───┴───┴───┴───┴───┘
Note that we cannot match ✅「⟨⋯ 0 ⋯ 3 ⋯ 1 ⋯⟩」 unless 0 = [4th]. So we have
┌───┬───┬───┬───┬───┬───┐
│5th│ 4■│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ │ │ │ 4 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ 0 │ │ │ │ 4 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 5 │ │ 3 │ │
└───┴───┴───┴───┴───┴───┘
We then consider where to place 5. By ⛔「⟨ ⁰ᵗʰ↑ ⟩ after ⟨→⟩」, we have 5 != [1st]; and by ⛔「⟨ ³ʳᵈa ¹ˢᵗb ⁰ᵗʰc ⟩, (abc)₁₀ ≥ 314」, we have 5 != [3rd]. If 5 = [2nd], then ✅「⟨⋯ 0 ⋯ 3 ⋯ 1 ⋯⟩」 gives
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 0 │ 3 │ 5 │ 1 │ 4 │
└───┴───┴───┴───┴───┴───┘
and we fail to avoid ⛔「⟨ ³ʳᵈa ¹ˢᵗb ⁰ᵗʰc ⟩, (abc)₁₀ ≥ 314」. Therefore, 5 = [5th]:
┌───┬───┬───┬───┬───┬───┐
│ 5■│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 0 │ │ │ │ 4 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 5 │ 0 │ │ │ │ 4 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ │ │ 3 │ │
└───┴───┴───┴───┴───┴───┘
And then we consider the position of 1. To match ✅「⟨⋯ 0 ⋯ 3 ⋯ 1 ⋯⟩」, we need 1 = [2nd] or [1st]. If it is [2nd], then the preceding pattern gives
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 5 │ 0 │ 3 │ 1 │ 2 │ 4 │
└───┴───┴───┴───┴───┴───┘
and we fail to avoid ⛔「⟨ ³ʳᵈa ¹ˢᵗb ⁰ᵗʰc ⟩, (abc)₁₀ ≥ 314」 again. Therefore, it has to be [1st]:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 5 │ 0 │ │ │ │ 4 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 5 │ 0 │ │ │ 1 │ 4 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ │ │ 3 │ │
└───┴───┴───┴───┴───┴───┘
Finally, using ⛔「⟨ ³ʳᵈa ¹ˢᵗb ⁰ᵗʰc ⟩, (abc)₁₀ ≥ 314」 once more, we finish by
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3■│ 2■│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 5 │ 0 │ │ │ 1 │ 4 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 5 │ 0 │ 2 │ │ 1 │ 4 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 5 │ 0 │ 2 │ 3 │ 1 │ 4 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.12