Rearrange the digits in ⟨125034⟩ to meet the rules below.
⟨ ⁵ᵗʰ▨ ⁴ᵗʰ▨ ³ʳᵈ▨ ²ⁿᵈ▨ ¹ˢᵗ▨ ⁰ᵗʰ▨ ⟩
✅Match
4th → 0|1|2
⟨ ▧ ▧ − − − ▢ ⟩, Σ▧ ≥ ▢
⟨ ⁵ᵗʰ↑ ⟩ after 3×⟨←⟩
Jump(0,2) = 0
⛔Avoid
⟨⋯ 2 ⋯ a ⋯⟩, a = 0|1
max {p4, p2, p0} = 5
#125034_v2.13
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ 1 │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ 3 │ 1 │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 3 │ 1 │ │ 4 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 3 │ 1 │ │ 4 │ 0 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 3 │ 1 │ │ 4 │ 0 │ 2 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 3 │ 1 │ 5 │ 4 │ 0 │ 2 │▒
└───┴───┴───┴───┴───┴───┘▒
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Proof of 2026-07-07 WR
══════════════════════
Notation: if nth -> a, then we write [nth] = a.
Firstly, we use ✅「4th → 0|1|2」 to determine [4th].
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ ▬ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
If it is 2, then we cannot avoid ⛔「⟨⋯ 2 ⋯ a ⋯⟩, a = 0|1」. Else if it is 0, then by ✅「Jump(0,2) = 0」 and the preceding pattern, we have
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 1 │ 0 │ 2 │ │ │ │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ 5 │ │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
We will fail to match ✅「⟨ ▧ ▧ − − − ▢ ⟩, Σ▧ ≥ ▢」, which is a contradiction. Therefore, [4th] has to be 1:
┌───┬───┬───┬───┬───┬───┐
│5th│ 4■│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ 1 │ │ │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ 5 │ 0 │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
Next, we consider what [5th] is. We see that:
• It is not 0 by ✅「⟨ ▧ ▧ − − − ▢ ⟩, Σ▧ ≥ ▢」.
• It is not 2 by ⛔「⟨⋯ 2 ⋯ a ⋯⟩, a = 0|1」.
• It is not 5 by ✅「⟨ ⁵ᵗʰ↑ ⟩ after 3×⟨←⟩」.
So, it has to be 3 or 4. If it is 4, then to match ✅「⟨ ⁵ᵗʰ↑ ⟩ after 3×⟨←⟩」, we need
┌───┬───┬───┬───┬───┬───┐
│ 5▲│4th│3rd│ 2▲│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 4 │ 1 │ │ 5 │ │ │
└───┴───┴───┴───┴───┴───┘
and we will fail to avoid ⛔「max {p4, p2, p0} = 5」. As a result, [5th] = 3:
┌───┬───┬───┬───┬───┬───┐
│ 5■│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 1 │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ 3 │ 1 │ │ │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ 5 │ 0 │ │ 4 │
└───┴───┴───┴───┴───┴───┘
To avoid ⛔「max {p4, p2, p0} = 5」, we need 5!=[2nd]. Therefore, to match ✅「⟨ ⁵ᵗʰ↑ ⟩ after 3×⟨←⟩」, we need [2nd] = 4:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│ 2■│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 3 │ 1 │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 3 │ 1 │ │ 4 │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ 5 │ 0 │ │ │
└───┴───┴───┴───┴───┴───┘
By ✅「Jump(0,2) = 0」, 0 and 2 are adjacent; and by ⛔「⟨⋯ 2 ⋯ a ⋯⟩, a = 0|1」, 2 is to the right of 0. There is only one way to do so:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3■│2nd│ 1■│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 3 │ 1 │ │ 4 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 3 │ 1 │ │ 4 │ 0 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 3 │ 1 │ │ 4 │ 0 │ 2 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 3 │ 1 │ 5 │ 4 │ 0 │ 2 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.13