2025-07-01 WR

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

⟨5th 4th 3rd 2nd 1st 0th⟩

✅Match

4th → a, 0th → b, ab=3+5n

⟨ ⁵ᵗʰa     ²ⁿᵈb     ⟩, max⟦a,b⟧ = 5

2nd → a, 1st → b, ab=5

⛔Avoid

⟦0,1⟧ ∋ 2

⟨⋯ 5 ⋯ 2 ⋯⟩

#125034_v2.9

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

Proof of 2025-07-01 WR
══════════════════════

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

By ✅「2nd → a, 1st → b, ab=5」, we have 

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

On the other hand, note that ✅「⟨ ⁵ᵗʰa     ²ⁿᵈb     ⟩, max⟦a,b⟧ = 5」 implies 5 != [1st]. Therefore, 5 = [2nd]:

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

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

Then (1) implies [1st] = 1:

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

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

Next, we consider ✅「4th → a, 0th → b, ab=3+5n」. To match it, we need

(2) {[4th], [0th]} = {2,4}.

To avoid ⛔「⟨⋯ 5 ⋯ 2 ⋯⟩」, we cannot have 2 = [0th]. So we have 2 = [4th] and 4 = [0th]:

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

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

Finally, to avoid ⛔「⟦0,1⟧ ∋ 2」, we finish by

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

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

Q.E.D.

#125034_v2.9