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2026-05-12 WR

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

⟨ ⁵ᵗʰ▨ ⁴ᵗʰ▨ ³ʳᵈ▨ ²ⁿᵈ▨ ¹ˢᵗ▨ ⁰ᵗʰ▨ ⟩

✅Match
⟨     ³ʳᵈa     ⁰ᵗʰb ⟩, max⟦a,b⟧ = 4
⟨⋯ Perm(0,2,5) ⋯⟩
5th → a, 0th → b, ab=3+5n
min ⊢5⊣ ≤ 1

⛔Avoid
3rd → a, 1st → b, |a-b|=3
4th|2nd|1st|0th → 2
⟨⋯ 0 ⋯ ? 1 ⋯ (?+1)⟩ (?≠1,0)

#125034_v2.13


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

Proof of 2026-05-12 WR
══════════════════════

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

✅「⟨     ³ʳᵈa     ⁰ᵗʰb ⟩, max⟦a,b⟧ = 4」 implies

(1) 5 = [5th] | [4th].

If it is [5th], then we cannot match ✅「5th → a, 0th → b, ab=3+5n」. Therefore, 5 = [4th].

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │   │ 0 │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘

Next, we consider where to place 4.

(2) We claim that indeed 4 = [1st]:

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

------------------------------

(3.1) It is proved by contradiction. Firstly, suppose 4 = [5th]:

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

Then ✅「5th → a, 0th → b, ab=3+5n」 implies [0th] = 2:

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

We do not match ✅「⟨⋯ Perm(0,2,5) ⋯⟩」, which is a contradiction.

(3.2) Else if 4 = [3rd]:

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

then we cannot match ✅「⟨⋯ Perm(0,2,5) ⋯⟩」.

(3.3) Else if 4 = [2nd]:

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

then to match ✅「⟨⋯ Perm(0,2,5) ⋯⟩」, we need {[5th], [3rd]} = {0,2}. But then we cannot match ✅「5th → a, 0th → b, ab=3+5n」.

(3.4) Lastly, if 4 = [0th]:

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

then using ✅「5th → a, 0th → b, ab=3+5n」 and ✅「⟨⋯ Perm(0,2,5) ⋯⟩」, we get

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

Two cases follow:

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

In the former case, we match ⛔「3rd → a, 1st → b, |a-b|=3」. In the latter case, we match ⛔「⟨⋯ 0 ⋯ ? 1 ⋯ (?+1)⟩ (?≠1,0)」. Both are contradictions.

------------------------------

We have verified (2). Accordingly, we get

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │   │ 0 │ 3 │   │
└───┴───┴───┴───┴───┴───┘

Then, to match ✅「5th → a, 0th → b, ab=3+5n」, we need

(4) {[5th], [0th]} = {1,3}.

It implies

(5) {[3rd], [2nd]} = {0,2}.

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

In view of ⛔「4th|2nd|1st|0th → 2」, we have

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

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

Finally, using ✅「min ⊢5⊣ ≤ 1」, we finish by

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

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

Q.E.D.

#125034_v2.13