2026-04-21 WR

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

⟨ ⁵ᵗʰ▨ ⁴ᵗʰ▨ ³ʳᵈ▨ ²ⁿᵈ▨ ¹ˢᵗ▨ ⁰ᵗʰ▨ ⟩

✅Match

⟨ ⁵ᵗʰ↑ ⁴ᵗʰ↓ ³ʳᵈ↓ ²ⁿᵈ↑ ¹ˢᵗ↓ ⁰ᵗʰ↑ ⟩ after 5−⟨⋯⟩

⟨ − − ▧ ▢ ▢ ▢ ⟩, ▧ ≥ Σ▢

4th → a, 2nd → b, a+b=6

⛔Avoid

3rd → a, 1st → b, a+b=7

#125034_v2.13

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

Proof of 2026-04-21 WR
══════════════════════

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

By ✅「⟨ ⁵ᵗʰ↑ ⁴ᵗʰ↓ ³ʳᵈ↓ ²ⁿᵈ↑ ¹ˢᵗ↓ ⁰ᵗʰ↑ ⟩ after 5−⟨⋯⟩」, we have

(1) 
{[5th], [2nd], [0th]} = {0,1,2} and
{[4th], [3rd], [1st]} = {3,4,5}.

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

We consider what [1st] is. By ✅「⟨ − − ▧ ▢ ▢ ▢ ⟩, ▧ ≥ Σ▢」, we have [3rd] > [1st]. Therefore, there are only two possibilities:

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

Note that we cannot match ✅「4th → a, 2nd → b, a+b=6」 if (2.1) holds. Therefore, (2.2) holds and we get

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 5 │ 0 │   │ 4 │
└───┴───┴───┴───┴───┴───┘

By (2.2), we have [3rd] = 4 or 5. If it is 4, then we match ⛔「3rd → a, 1st → b, a+b=7」, which is a contradiction. Therefore, it is 5, and [4th] = 4 follows:

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

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

Using ✅「4th → a, 2nd → b, a+b=6」, we then have

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

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

Finally, in view of ✅「⟨ − − ▧ ▢ ▢ ▢ ⟩, ▧ ≥ Σ▢」, we finish by

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

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

Q.E.D.

#125034_v2.13