Rearrange the digits in ⟨125034⟩ to meet the rules below.
⟨ ⁵ᵗʰ▨ ⁴ᵗʰ▨ ³ʳᵈ▨ ²ⁿᵈ▨ ¹ˢᵗ▨ ⁰ᵗʰ▨ ⟩
✅Match
4th → a, 2nd → b, |a-b|=2
⟨⋯ ᵃb ⋯⟩, a+b=8
0th → 0|5
4th → a, 1st → b, ab=12
⛔Avoid
⟦0,5⟧ ∋ 2,3
#125034_v2.14
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ 3 │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ 3 │ │ │ 4 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ 3 │ │ 1 │ 4 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ │ 3 │ 5 │ 1 │ 4 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ │ 3 │ 5 │ 1 │ 4 │ 0 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 2 │ 3 │ 5 │ 1 │ 4 │ 0 │▒
└───┴───┴───┴───┴───┴───┘▒
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Proof of 2026-08-25 WR
══════════════════════
Notation: if nth -> a, then we write [nth] = a.
By ✅「4th → a, 1st → b, ab=12」, we have
(1) {[4th], [3rd]} = {3,4}.
(1.1) We claim that indeed ([4th], [3rd]) = (3,4).
------------------------------
Suppose on the contrary (1.1) does not hold. So
┌───┬───┬───┬───┬───┬───┐
│5th│ 4▲│3rd│2nd│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 4 │ │ │ 3 │ │
└───┴───┴───┴───┴───┴───┘
To match ✅「4th → a, 2nd → b, |a-b|=2」, we need [2nd] = 2:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│ 2▲│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 4 │ │ 2 │ 3 │ │
└───┴───┴───┴───┴───┴───┘
By ✅「0th → 0|5」, we have [0th] = 0 or 5:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 4 │ │ 2 │ 3 │0 5│
└───┴───┴───┴───┴───┴───┘
No matter which happens, we cannot avoid ⛔「⟦0,5⟧ ∋ 2,3」. This shows a contradiction.
------------------------------
We have verified (1.1). Accordingly, we get
┌───┬───┬───┬───┬───┬───┐
│5th│ 4■│3rd│2nd│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ 3 │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ 3 │ │ │ 4 │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 5 │ 0 │ │ │
└───┴───┴───┴───┴───┴───┘
By ✅「4th → a, 2nd → b, |a-b|=2」, we have [2nd] = 1 or 5. If it is 5:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│ 2▲│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 3 │ │ 5 │ 4 │ │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ │ 0 │ │ │
└───┴───┴───┴───┴───┴───┘
then we will fail to match ✅「⟨⋯ ᵃb ⋯⟩, a+b=8」. Therefore, [2nd] is 1, not 5:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│ 2■│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 3 │ │ │ 4 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ 3 │ │ 1 │ 4 │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ 5 │ 0 │ │ │
└───┴───┴───┴───┴───┴───┘
Using ✅「⟨⋯ ᵃb ⋯⟩, a+b=8」, we have [3rd] = 5:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3■│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 3 │ │ 1 │ 4 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ │ 3 │ 5 │ 1 │ 4 │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ │ 0 │ │ │
└───┴───┴───┴───┴───┴───┘
Finally, in view of ✅「0th → 0|5」, we finish by
┌───┬───┬───┬───┬───┬───┐
│ 5■│4th│3rd│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 3 │ 5 │ 1 │ 4 │ │▒
├───┼───┼───┼───┼───┼───┤▒
│ │ 3 │ 5 │ 1 │ 4 │ 0 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 2 │ 3 │ 5 │ 1 │ 4 │ 0 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.14