Rearrange the digits in ⟨125034⟩ to meet the rules below.
⟨ ⁵ᵗʰ▨ ⁴ᵗʰ▨ ³ʳᵈ▨ ²ⁿᵈ▨ ¹ˢᵗ▨ ⁰ᵗʰ▨ ⟩
✅Match
Jump(1,4) = 1
4th → a, 0th → b, a+b=1+4n
2nd → a, 1st → b, ab=0+4n
⟨⋯ Perm(1,5) ⋯⟩
⛔Avoid
⟨⋯ 2 ⋯ 0 ⋯ 5 ⋯⟩
⟨⋯ ᵃb ⋯⟩, ab=4
⟨? ⋯ 2 ⋯ (?+1) ⋯⟩ (?≠2,1)
#125034_v2.13
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ 5 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ │ │ 5 │ 0 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ │ 1 │ 5 │ 0 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 4 │ │ 1 │ 5 │ 0 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 4 │ 3 │ 1 │ 5 │ 0 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 4 │ 3 │ 1 │ 5 │ 0 │ 2 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
Proof of 2026-06-09 WR
══════════════════════
Notation: if nth -> a, then we write [nth] = a.
We first show that
(1) 4 is not [2nd] or [1st].
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ │ / │ / │ │
└───┴───┴───┴───┴───┴───┘
------------------------------
Plainly, ⛔「⟨⋯ ᵃb ⋯⟩, ab=4」 implies 4 != [1st]. So, suppose 4 = [2nd]:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│ 2▲│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ │ 4 │ │ │
└───┴───┴───┴───┴───┴───┘
To match ✅「Jump(1,4) = 1」 and avoid ⛔「⟨⋯ ᵃb ⋯⟩, ab=4」 at the same time, we need 1 = [0th]:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ │ 4 │ │ 1 │
└───┴───┴───┴───┴───┴───┘
Then, ✅「⟨⋯ Perm(1,5) ⋯⟩」 implies [1st] = 5:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ │ 4 │ 5 │ 1 │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ │ 0 │ 3 │ │
└───┴───┴───┴───┴───┴───┘
To avoid ⛔「⟨? ⋯ 2 ⋯ (?+1) ⋯⟩ (?≠2,1)」, [5th] cannot be 0 or 3. Therefore, [5th] = 2:
┌───┬───┬───┬───┬───┬───┐
│ 5▲│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ │ │ 4 │ 5 │ 1 │
└───┴───┴───┴───┴───┴───┘
We cannot avoid ⛔「⟨⋯ 2 ⋯ 0 ⋯ 5 ⋯⟩」 now, which is a contradiction.
------------------------------
We have verified (1). Next, we show that
(2) 4 is not [4th] or [0th].
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ / │ │ │ │ / │
└───┴───┴───┴───┴───┴───┘
------------------------------
For, if it is not true, then there are two cases:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 4 │ │ │ │ │
├───┼───┼───┼───┼───┼───┤
│ │ │ │ │ │ 4 │
└───┴───┴───┴───┴───┴───┘
and ✅「Jump(1,4) = 1」 implies
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 4 │ │ 1 │ │ │
├───┼───┼───┼───┼───┼───┤
│ │ │ │ 1 │ │ 4 │
└───┴───┴───┴───┴───┴───┘
By ✅「⟨⋯ Perm(1,5) ⋯⟩」, 5 is adjacent to 1. As a result, we cannot match ✅「4th → a, 0th → b, a+b=1+4n」. This shows a contradiction.
------------------------------
We have verified (2). Combining (2) with (1), we get
(3) 4 = [5th] or [3rd].
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ X │ │ X │ │ │ │
└───┴───┴───┴───┴───┴───┘
Then, by combining (3) with ✅「2nd → a, 1st → b, ab=0+4n」, we have
(4) 0 is [2nd] or [1st].
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ X │ │ X │ Y │ Y │ │
└───┴───┴───┴───┴───┴───┘
We show that the same holds for 5:
(5) 5 is [2nd] or [1st].
------------------------------
Suppose on the contrary this is not true, then using (3), (4) and ✅「Jump(1,4) = 1」, there are four cases:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
(5.1) │ 5 │ │ 4 │ 0 │ 1 │ │
├───┼───┼───┼───┼───┼───┤
(5.2) │ X │ 5 │ X │ Y │ Y │ │
├───┼───┼───┼───┼───┼───┤
(5.3) │ 4 │ │ 5 │ Y │ Y │ │
├───┼───┼───┼───┼───┼───┤
(5.4) │ X │ │ X │ Y │ Y │ 5 │
└───┴───┴───┴───┴───┴───┘
They are contradictions because:
• if (5.1) holds, then we cannot match ✅「⟨⋯ Perm(1,5) ⋯⟩」;
• if (5.3) holds, then we cannot match ✅「Jump(1,4) = 1」;
• if (5.2) or (5.4) hold, then we cannot match ✅「4th → a, 0th → b, a+b=1+4n」, (3) and (4) at the same time.
------------------------------
We have verified (5). Combining (5) with (4), we get
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ X │ │ X │0 5│0 5│ │
└───┴───┴───┴───┴───┴───┘
Using ✅「Jump(1,4) = 1」, we further obtain
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│1 4│2 3│1 4│0 5│0 5│2 3│
└───┴───┴───┴───┴───┴───┘
Note that to match ✅「⟨⋯ Perm(1,5) ⋯⟩」, we need to place 0,5 as follows:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│ 2■│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ 5 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ │ │ 5 │ 0 │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ │ │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
and place 1,4 this way:
┌───┬───┬───┬───┬───┬───┐
│ 5■│4th│ 3■│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ │ 5 │ 0 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ │ 1 │ 5 │ 0 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 4 │ │ 1 │ 5 │ 0 │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ │ │ 3 │ │
└───┴───┴───┴───┴───┴───┘
Finally, to avoid ⛔「⟨? ⋯ 2 ⋯ (?+1) ⋯⟩ (?≠2,1)」, we finish by
┌───┬───┬───┬───┬───┬───┐
│5th│ 4■│3rd│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 4 │ │ 1 │ 5 │ 0 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 4 │ 3 │ 1 │ 5 │ 0 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 4 │ 3 │ 1 │ 5 │ 0 │ 2 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.13