2026-02-17 WR

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

⟨ ⁵ᵗʰ▨ ⁴ᵗʰ▨ ³ʳᵈ▨ ²ⁿᵈ▨ ¹ˢᵗ▨ ⁰ᵗʰ▨ ⟩

✅Match

⟨⋯ ᵃb ⋯⟩, |a-b|=0

⟨⋯ 3 ⋯ 4 ⋯⟩

min ⊢3⊣ = 2

⛔Avoid

⟨⋯ 2 ⋯ a ⋯⟩, a = 1|3

5th → a, 1st → b, a+b=0+5n

#125034_v2.12

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

Proof of 2026-02-17 WR
══════════════════════

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

By ✅「min ⊢3⊣ = 2」, we see that 2 and 3 are adjacent. Combining this with ⛔「⟨⋯ 2 ⋯ a ⋯⟩, a = 1|3」, we have to match the following pattern:

(1) ⟨⋯ 1 ⋯ 3 2 ⋯⟩.

To match ✅「min ⊢3⊣ = 2」, the digit adjacent to and to the left of 3 can only be 4 or 5. In view of ✅「⟨⋯ 3 ⋯ 4 ⋯⟩」, it is 5 indeed. Accordingly, we have to match

(2) ⟨⋯ 1 ⋯ 5 3 2 ⋯ 4 ⋯⟩.

There are two ways to match (2):

      ┌───┬───┬───┬───┬───┬───┐
      │5th│4th│3rd│2nd│1st│0th│
      ╞═══╪═══╪═══╪═══╪═══╪═══╡
(3.1) │   │   │ 5 │ 3 │ 2 │ 4 │
      ├───┼───┼───┼───┼───┼───┤
(3.2) │ 1 │ 5 │ 3 │ 2 │   │   │
      └───┴───┴───┴───┴───┴───┘

Note that if (3.1) holds, then we cannot match ✅「⟨⋯ ᵃb ⋯⟩, |a-b|=0」. Therefore, (3.2) holds.

Finally, to avoid ⛔「5th → a, 1st → b, a+b=0+5n」, we finish by ⟨153204⟩.

Q.E.D.

#125034_v2.12