2023-12-07 Q1(m=6)

Rearrange the digits in ⟨1263045⟩ to meet the rules below.

⟨6th 5th 4th 3rd 2nd 1st 0th⟩

✅Match

⟨⋯ Perm(1,4) ⋯⟩

⟨⋯ 5 ⋯ 6 ⋯⟩

⟦0,4⟧ ∋ 5,6

1st → 0

⟨? ⋯ 3 (?+1) ⋯ 5 ⋯⟩ (?≠3,2)

#125034_v2.2

       ┌───┬───┬───┬───┬───┬───┬───┐
       │6th│5th│4th│3rd│2nd│1st│0th│▒
       ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │   │   │   │   │   │ 0 │   │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │   │   │   │   │   │ 0 │ 2 │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ 4 │   │   │   │   │ 0 │ 2 │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 4 │ 1 │   │   │   │ 0 │ 2 │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 4 │ 1 │ 3 │   │   │ 0 │ 2 │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 4 │ 1 │ 3 │ 5 │   │ 0 │ 2 │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 4 │ 1 │ 3 │ 5 │ 6 │ 0 │ 2 │▒
       └───┴───┴───┴───┴───┴───┴───┘▒
        ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒

Proof of 2023-12-07 Q1(m=6)
═══════════════════════════

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

Plainly, our first step follows from ✅「1st → 0」. Combining ✅「⟨⋯ Perm(1,4) ⋯⟩」, ✅「⟨⋯ 5 ⋯ 6 ⋯⟩」, and ✅「⟦0,4⟧ ∋ 5,6」, we need to match the following patterns:

(1) ⟨⋯ 4 ⋯ 5 ⋯ 6 ⋯ 0 ⋯⟩

(2) ⟨⋯ 1 ⋯ 5 ⋯ 6 ⋯ 0 ⋯⟩

By 0 = [1st], (1) and (2), we see that [0th] != 0,1,4,5,6. Hence [0th] = 2 or 3. Noting that ✅「⟨? ⋯ 3 (?+1) ⋯ 5 ⋯⟩ (?≠3,2)」 implies 3 is not in corners, we have [0th] = 2.

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │   │ 6 │ 3 │   │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘

(3) Next, we determine the value of [6th]. By ✅「⟨? ⋯ 3 (?+1) ⋯ 5 ⋯⟩ (?≠3,2)」, we have [6th] = 1 or 4.

It is not possible that [6th] = 1, for otherwise the preceding pattern becomes

 ⟨1 ⋯ 32 ⋯ 5 ⋯⟩

which is a contradiction as we already have 2 = [0th], so that no number can be at the right of 2.

Therefore, back to (3), we have [6th] = 4, and applying ✅「⟨⋯ Perm(1,4) ⋯⟩」, we get

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│   │   │ 6 │ 3 │   │   │ 5 │
└───┴───┴───┴───┴───┴───┴───┘

Now ✅「⟨? ⋯ 3 (?+1) ⋯ 5 ⋯⟩ (?≠3,2)」 becomes:

    ⟨4 ⋯ 35 ⋯⟩

Combining this pattern with (1), we reach

       ┌───┬───┬───┬───┬───┬───┬───┐
       │6th│5th│ 4■│ 3■│ 2■│1st│0th│▒
       ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
       │ 4 │ 1 │   │   │   │ 0 │ 2 │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 4 │ 1 │ 3 │   │   │ 0 │ 2 │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 4 │ 1 │ 3 │ 5 │   │ 0 │ 2 │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 4 │ 1 │ 3 │ 5 │ 6 │ 0 │ 2 │▒
       └───┴───┴───┴───┴───┴───┴───┘▒
        ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒

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

Q.E.D.

#125034_v2.2