2026-06-30 WR

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

⟨ ⁵ᵗʰ▨ ⁴ᵗʰ▨ ³ʳᵈ▨ ²ⁿᵈ▨ ¹ˢᵗ▨ ⁰ᵗʰ▨ ⟩

✅Match

⟨⋯ 1 ⋯ 4 ⋯ 3 ⋯⟩

median {p5, p3, p1} = 1

⟨ − − ▧ − ▢ ▢ ⟩, ▧ ≥ Σ▢

⟨   ⁴ᵗʰb ³ʳᵈc ²ⁿᵈa     ⟩, a > b > c

#125034_v2.13

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

Proof of 2026-06-30 WR
══════════════════════

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

By ✅「median {p5, p3, p1} = 1」, we have

(1) 1 ∈ {[5th], [3rd], [1st]}, and

(2) 0 ∈ {[5th], [3rd], [1st]}.

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

We consider how to match (1). If 1=[1st], then we cannot match ✅「⟨⋯ 1 ⋯ 4 ⋯ 3 ⋯⟩」. Else if 1=[3rd], then we cannot match ✅「⟨ − − ▧ − ▢ ▢ ⟩, ▧ ≥ Σ▢」. Therefore, 1=[5th]:

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│   │ 2 │ 5 │ 0 │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘

Then, to match (2) and ✅「⟨ − − ▧ − ▢ ▢ ⟩, ▧ ≥ Σ▢」 at the same time, 0 has to be [1st]:

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│   │ 2 │ 5 │   │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘

Next, we consider what [3rd] is. To match ✅「⟨   ⁴ᵗʰb ³ʳᵈc ²ⁿᵈa     ⟩, a > b > c」, it can only be 2 or 3. If it is 2, then we cannot match ✅「⟨ − − ▧ − ▢ ▢ ⟩, ▧ ≥ Σ▢」. Therefore, it is 3:

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

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

Then ✅「⟨⋯ 1 ⋯ 4 ⋯ 3 ⋯⟩」 gives 4=[4th]:

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

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

Finally, to match ✅「⟨   ⁴ᵗʰb ³ʳᵈc ²ⁿᵈa     ⟩, a > b > c」, we finish by

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

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

Q.E.D.

#125034_v2.13