2025-10-28 WR

Rearrange the digits in ⟨125034⟩ to meet the rules below.

⟨5th 4th 3rd 2nd 1st 0th⟩

✅Match

⟨⋯ ? ⋯ 1 ⋯ (?+1)⟩ (?≠1,0)

3rd → a, 1st → b, ab=2+5n

⟨⋯ Perm(0,2,3) ⋯⟩

⛔Avoid

max ⊢0⊣ ≤ 4

#125034_v2.10

       ┌───┬───┬───┬───┬───┬───┐
       │5th│4th│3rd│2nd│1st│0th│▒
       ╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │   │ 0 │   │   │   │   │▒
       ├───┼───┼───┼───┼───┼───┤▒
Step 2 │ 5 │ 0 │   │   │   │   │▒
       ├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 5 │ 0 │   │   │ 1 │   │▒
       ├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 5 │ 0 │   │   │ 1 │ 4 │▒
       ├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 5 │ 0 │ 2 │   │ 1 │ 4 │▒
       ├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 5 │ 0 │ 2 │ 3 │ 1 │ 4 │▒
       └───┴───┴───┴───┴───┴───┘▒
        ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒

Proof of 2025-10-28 WR
══════════════════════

Notation: if nth -> a, then we write [nth] = a.

By combining ✅「⟨⋯ Perm(0,2,3) ⋯⟩」 with ⛔「max ⊢0⊣ ≤ 4」, we see that one of the following holds:

(1) ⟨⋯ 5 0 Perm(2,3) ⋯⟩ or ⟨⋯ Perm(2,3) 0 5 ⋯⟩.

There are six possible ways to do so:

      ┌───┬───┬───┬───┬───┬───┐
      │5th│4th│3rd│2nd│1st│0th│
      ╞═══╪═══╪═══╪═══╪═══╪═══╡
(2.1) │   │   │ ▬ │ ▬ │ 0 │ 5 │
      ├───┼───┼───┼───┼───┼───┤
(2.2) │   │   │ 5 │ 0 │ ▬ │ ▬ │
      ├───┼───┼───┼───┼───┼───┤
(2.3) │   │ ▬ │ ▬ │ 0 │ 5 │   │
      ├───┼───┼───┼───┼───┼───┤
(2.4) │   │ 5 │ 0 │ ▬ │ ▬ │   │
      ├───┼───┼───┼───┼───┼───┤
(2.5) │ ▬ │ ▬ │ 0 │ 5 │   │   │
      ├───┼───┼───┼───┼───┼───┤
(2.6) │ 5 │ 0 │ ▬ │ ▬ │   │   │
      └───┴───┴───┴───┴───┴───┘

Note that to match ✅「3rd → a, 1st → b, ab=2+5n」, [3rd] cannot be 0 or 5, and the same is true for [1st]. Accordingly, only (2.6) is possible. We get

       ┌───┬───┬───┬───┬───┬───┐
       │ 5■│ 4■│3rd│2nd│1st│0th│▒
       ╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │   │ 0 │   │   │   │   │▒
       ├───┼───┼───┼───┼───┼───┤▒
Step 2 │ 5 │ 0 │   │   │   │   │▒
       └───┴───┴───┴───┴───┴───┘▒
        ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒

--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │   │   │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘

We also have {[3rd], [2nd]} = {2,3}, and {[1st], [0th]} = {1,4} follows:

┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 5 │ 0 │2 3│2 3│1 4│1 4│
└───┴───┴───┴───┴───┴───┘

By ✅「⟨⋯ ? ⋯ 1 ⋯ (?+1)⟩ (?≠1,0)」, 1 is not in the right corner (0th). Therefore, [1st] = 1 and [0th] = 4.

       ┌───┬───┬───┬───┬───┬───┐
       │5th│4th│3rd│2nd│ 1■│ 0■│▒
       ╞═══╪═══╪═══╪═══╪═══╪═══╡▒
       │ 5 │ 0 │   │   │   │   │▒
       ├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 5 │ 0 │   │   │ 1 │   │▒
       ├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 5 │ 0 │   │   │ 1 │ 4 │▒
       └───┴───┴───┴───┴───┴───┘▒
        ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒

--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│   │ 2 │   │   │ 3 │   │
└───┴───┴───┴───┴───┴───┘

Finally, using ✅「3rd → a, 1st → b, ab=2+5n」, 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_v2.10