Rearrange the digits in ⟨125034⟩ to meet the rules below.
⟨ ⁵ᵗʰ▨ ⁴ᵗʰ▨ ³ʳᵈ▨ ²ⁿᵈ▨ ¹ˢᵗ▨ ⁰ᵗʰ▨ ⟩
✅Match
{p2, p1, p0} = ? + {0,1,3}
⛔Avoid
max {p5, p1, p0} = 4
⟨⋯ Perm(1,2,4) ⋯⟩
⟨ − ▧ ▢ ▢ − − ⟩, ▧ ≤ Σ▢
⟨⋯ ? ⋯ 0 ⋯ (?−1)⟩ (?≠1)
Jump(2,3) = 0
#125034_v2.15
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ │ 5 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ 4 │ │ │ 5 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ 4 │ │ 3 │ 5 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ │ 4 │ │ 3 │ 5 │ 2 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 1 │ 4 │ │ 3 │ 5 │ 2 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 1 │ 4 │ 0 │ 3 │ 5 │ 2 │▒
└───┴───┴───┴───┴───┴───┘▒
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Proof of 2026-11-10 WR
══════════════════════
Notation: if nth -> a, then we write [nth] = a.
By ✅「{p2, p1, p0} = ? + {0,1,3}」, we have
(1) {p2, p1, p0} = {0,1,3} or {1,2,4} or {2,3,5}.
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ │ ▬ │ ▬ │ ▬ │
└───┴───┴───┴───┴───┴───┘
(1.1) We first show that it is not {0,1,3}.
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For, suppose it is {0,1,3}:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│245│245│245│013│013│013│
└───┴───┴───┴───┴───┴───┘
Then, to avoid ⛔「⟨⋯ ? ⋯ 0 ⋯ (?−1)⟩ (?≠1)」, [0th] has to be 0:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│245│245│245│1 3│1 3│ 0 │
└───┴───┴───┴───┴───┴───┘
Consider where to place 4. To avoid ⛔「max {p5, p1, p0} = 4」 and ⛔「⟨ − ▧ ▢ ▢ − − ⟩, ▧ ≤ Σ▢」, we need 4 = [4th]:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│2 5│ 4 │2 5│1 3│1 3│ 0 │
└───┴───┴───┴───┴───┴───┘
Using ⛔「⟨ − ▧ ▢ ▢ − − ⟩, ▧ ≤ Σ▢」 again, we get
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 5 │ 4 │ 2 │1 3│1 3│ 0 │
└───┴───┴───┴───┴───┴───┘
But then no matter what [2nd] is, we will match ⛔「⟨⋯ Perm(1,2,4) ⋯⟩」 or ⛔「Jump(2,3) = 0」. This shows a contradiction.
------------------------------
We have verified (1.1). As a result
(2) {p2, p1, p0} = {1,2,4} or {2,3,5}.
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ │ ▬ │ ▬ │ ▬ │
└───┴───┴───┴───┴───┴───┘
It cannot be {1,2,4} in view of ⛔「⟨⋯ Perm(1,2,4) ⋯⟩」. Therefore,
(3) {p2, p1, p0} = {2,3,5}.
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│014│014│014│235│235│235│
└───┴───┴───┴───┴───┴───┘
Given that, to avoid ⛔「Jump(2,3) = 0」, [1st] has to be 5:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │014│014│014│2 3│ 5 │2 3│▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
------------------------------
Note that if [4th]!=4, then [4th] = 0 or 1, and we cannot avoid ⛔「⟨ − ▧ ▢ ▢ − − ⟩, ▧ ≤ Σ▢」. So, [4th] has to be 4:
┌───┬───┬───┬───┬───┬───┐
│5th│ 4■│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 2 │0 1│ 4 │0 1│2 3│ 5 │2 3│▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
------------------------------
We consider whether ([2nd], [0th]) = (2,3) or (3,2). If it is (2,3):
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│0 1│ 4 │0 1│ 2 │ 5 │ 3 │
└───┴───┴───┴───┴───┴───┘
then no matter what [3rd] is, we will match ⛔「⟨⋯ Perm(1,2,4) ⋯⟩」 or ⛔「⟨⋯ ? ⋯ 0 ⋯ (?−1)⟩ (?≠1)」. So, we have ([2nd], [0th]) = (3,2) instead:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│ 2■│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 3 │0 1│ 4 │0 1│ 3 │ 5 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │0 1│ 4 │0 1│ 3 │ 5 │ 2 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
------------------------------
Finally, to avoid ⛔「⟨ − ▧ ▢ ▢ − − ⟩, ▧ ≤ Σ▢」, we finish by
┌───┬───┬───┬───┬───┬───┐
│ 5■│4th│ 3■│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 5 │ 1 │ 4 │ │ 3 │ 5 │ 2 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 1 │ 4 │ 0 │ 3 │ 5 │ 2 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
------------------------------
Q.E.D.
#125034_v2.15