2026-10-27 WR

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

⟨ ⁵ᵗʰ▨ ⁴ᵗʰ▨ ³ʳᵈ▨ ²ⁿᵈ▨ ¹ˢᵗ▨ ⁰ᵗʰ▨ ⟩

✅Match

⟨     ³ʳᵈa   ¹ˢᵗb   ⟩, min⟦a,b⟧ = 2

⟨⋯ 3 ⋯ 2 ⋯⟩

2nd|0th → 0

⟨⋯ ᵃb ⋯⟩, ab=2+7n

⛔Avoid

median {p4, p2, p0} = 1

Jump(3,4) = 1

⟨⋯ 5 ⋯ ? ⋯ 2 (?−3)⟩ (?≠5)

{p4, p2, p1} = ? + {0,2,3}

#125034_v2.15

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

Proof of 2026-10-27 WR
══════════════════════

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

By ✅「⟨     ³ʳᵈa   ¹ˢᵗb   ⟩, min⟦a,b⟧ = 2」, we have

(1) 0,1 ∈ {[5th], [4th], [0th]}.

Combining this with ✅「2nd|0th → 0」, we see that 0 = [0th]:

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 5 │   │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘

Then (1) implies 1 = [5th] or [4th]. If it is [4th], then we would match ⛔「median {p4, p2, p0} = 1」, which is not allowed. Therefore, 1 = [5th]:

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

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

Next, we consider ✅「⟨⋯ ᵃb ⋯⟩, ab=2+7n」. To match this pattern, the product "ab" can only be 2 or 9, and we need

(2) [1st] = 2 or [3rd] = 3.

Note that if [3rd] = 3, then we would match ⛔「{p4, p2, p1} = ? + {0,2,3}」, which is not allowed. Therefore

(3) [3rd] ≠ 3.

Combining this with (2), we have [1st] = 2:

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

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

By (3), we have 3 = [4th] or [2nd]. Given this, to avoid ⛔「Jump(3,4) = 1」, we need 4 = [3rd]:

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

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

Finally, to avoid ⛔「⟨⋯ 5 ⋯ ? ⋯ 2 (?−3)⟩ (?≠5)」, we finish by

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

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

Q.E.D.

#125034_v2.15