2026-06-23 WR

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

⟨ ⁵ᵗʰ▨ ⁴ᵗʰ▨ ³ʳᵈ▨ ²ⁿᵈ▨ ¹ˢᵗ▨ ⁰ᵗʰ▨ ⟩

✅Match

5th → a, 1st → b, a+b=9

Jump(3,5) ≥ 2

⛔Avoid

⟨ − ▧ ▢ ▢ − − ⟩, ▧ ≤ Σ▢

5th|2nd|1st|0th → 1

min {p5, p4} = 2

#125034_v2.13

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

Proof of 2026-06-23 WR
══════════════════════

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

By ⛔「5th|2nd|1st|0th → 1」, we have

(1) 1 = [4th] or [3rd].

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

If it is [4th], then we cannot avoid ⛔「⟨ − ▧ ▢ ▢ − − ⟩, ▧ ≤ Σ▢」. Therefore, 1 = [3rd]:

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

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

To continue, note that ✅「5th → a, 1st → b, a+b=9」 implies

(2) {[5th], [1st]} = {4,5}.

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

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

We determine the position of 2. By ⛔「min {p5, p4} = 2」, 2 != [4th], and by ⛔「⟨ − ▧ ▢ ▢ − − ⟩, ▧ ≤ Σ▢」, 2 != [2nd]. Therefore, 2 = [0th]:

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

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

Now, we have 3 = [4th] or [2nd]. If it is [2nd], then we cannot avoid ⛔「⟨ − ▧ ▢ ▢ − − ⟩, ▧ ≤ Σ▢」. Therefore, 3 = [4th]:

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

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

Finally, using ✅「Jump(3,5) ≥ 2」, we finish by

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

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

Q.E.D.

#125034_v2.13