2026-02-24 WR

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

⟨ ⁵ᵗʰ▨ ⁴ᵗʰ▨ ³ʳᵈ▨ ²ⁿᵈ▨ ¹ˢᵗ▨ ⁰ᵗʰ▨ ⟩

✅Match

⟨? 4 ⋯ (?−4) ⋯ 0 ⋯⟩ (?≠4)

⟨⋯ ? ⋯ 2 ⋯ (?−2)⟩ (?≠2,4)

⛔Avoid

min ⊢1⊣ = 1

⟨⋯ 2 ⋯ ? 0 ⋯ (?+2)⟩ (?≠0)

#125034_v2.12

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

Proof of 2026-02-24 WR
══════════════════════

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

By ✅「⟨? 4 ⋯ (?−4) ⋯ 0 ⋯⟩ (?≠4)」, plainly we have [4th] = 4:

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 5 │ 0 │ 3 │   │
└───┴───┴───┴───┴───┴───┘

Also, ✅「⟨? 4 ⋯ (?−4) ⋯ 0 ⋯⟩ (?≠4)」 implies that the leftmost digit is at least 4. Therefore, it is 5:

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │   │ 0 │ 3 │   │
└───┴───┴───┴───┴───┴───┘

Next, we consider the rightmost digit. To match ✅「⟨⋯ ? ⋯ 2 ⋯ (?−2)⟩ (?≠2,4)」, it can only be 1 or 3. If it is 1, then we cannot match ✅「⟨? 4 ⋯ (?−4) ⋯ 0 ⋯⟩ (?≠4)」. Therefore, it is 3:

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │   │ 0 │   │   │
└───┴───┴───┴───┴───┴───┘

Note that to avoid ⛔「min ⊢1⊣ = 1」, we need 0 and 1 to be adjacent. Combining this with ✅「⟨? 4 ⋯ (?−4) ⋯ 0 ⋯⟩ (?≠4)」, we have to match the following pattern:

(1) ⟨⋯ 1 0 ⋯⟩.

To match (1) and avoid ⛔「⟨⋯ 2 ⋯ ? 0 ⋯ (?+2)⟩ (?≠0)」 at the same time, 2 has to be at the 1st position:

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │   │   │ 0 │   │   │
└───┴───┴───┴───┴───┴───┘

Therefore, we finish by

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

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

Q.E.D.

#125034_v2.12