2026-07-21 WR

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

⟨ ⁵ᵗʰ▨ ⁴ᵗʰ▨ ³ʳᵈ▨ ²ⁿᵈ▨ ¹ˢᵗ▨ ⁰ᵗʰ▨ ⟩

✅Match

⟨⋯ Perm(1,2,3) ⋯⟩

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

⛔Avoid

⟨⋯ Perm(0,4) ⋯⟩

⟨⋯ a ⋯ 1 ⋯⟩, a = 0|5

{p5, p3, p2} = ? + {0,1,3}

#125034_v2.13

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

Proof of 2026-07-21 WR
══════════════════════

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

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

{[5th], [4th], [2nd]} = {0,1,2}

and

{[3rd], [1st], [0th]} = {3,4,5}.

    ┌───┬───┬───┬───┬───┬───┐
    │5th│4th│3rd│2nd│1st│0th│
    ╞═══╪═══╪═══╪═══╪═══╪═══╡
(1) │012│012│   │012│   │   │
    └───┴───┴───┴───┴───┴───┘

So there are only two ways to match ✅「⟨⋯ Perm(1,2,3) ⋯⟩」:

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

No matter which happens, we have 3 = [3rd]:

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

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

As either (2) or (3) holds, using (1) we also have

(4) 0 = [5th] or [2nd].

By ⛔「⟨⋯ a ⋯ 1 ⋯⟩, a = 0|5」, 0 is to the right of 1, so 0 is not [5th]. Therefore, 0 = [2nd]:

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

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

It implies (2) holds and we have

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

If 1 = [5th], then we will match ⛔「{p5, p3, p2} = ? + {0,1,3}」, which is a contradiction. Therefore, 1 = [4th] and 2 = [5th]:

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

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

Finally, to avoid ⛔「⟨⋯ Perm(0,4) ⋯⟩」, we finish by

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

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

Q.E.D.

#125034_v2.13