Web link

2026-06-02 WR

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

⟨ ⁵ᵗʰ▨ ⁴ᵗʰ▨ ³ʳᵈ▨ ²ⁿᵈ▨ ¹ˢᵗ▨ ⁰ᵗʰ▨ ⟩

✅Match
{p3, p2, p1} = ? + {0,1,3}
⟨⋯ ᵃb ⋯⟩, ab=4

⛔Avoid
⟦3,5⟧ ∋ 1,4
max ⊢2⊣ = 3
Jump(4,5) = 0
⟨         ¹ˢᵗa ⁰ᵗʰb ⟩, (ab)₁₀ ≤ 40
⟨ ⁵ᵗʰ↓ ⁴ᵗʰ↑ ³ʳᵈ↓ ²ⁿᵈ↑ ¹ˢᵗ↓ ⁰ᵗʰ↑ ⟩ after ⟨⇌⟩

#125034_v2.13


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

Proof of 2026-06-02 WR
══════════════════════

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

By ⛔「⟨         ¹ˢᵗa ⁰ᵗʰb ⟩, (ab)₁₀ ≤ 40」, we have

(1)  [1st] = 4 or 5.

(1.1) We claim that [1st] = 5.

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

For, suppose on the contrary [1st] = 4:

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

Then, ✅「{p3, p2, p1} = ? + {0,1,3}」 implies {[3rd], [2nd]} = {1,2}:

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

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

Consider what [0th] is. By ⛔「Jump(4,5) = 0」 it is not 5, and by ⛔「⟨         ¹ˢᵗa ⁰ᵗʰb ⟩, (ab)₁₀ ≤ 40」 it is not 0. Therefore, it is 3:

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

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

Note that we cannot avoid ⛔「⟦3,5⟧ ∋ 1,4」 now. This shows a contradiction.

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

We have verified (1.1). Accordingly, we get

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

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

Combining this with ✅「{p3, p2, p1} = ? + {0,1,3}」, we have

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

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

(2.1) We claim that ([3rd], [2nd]) = (3,2) actually.

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

For, if on the contrary ([3rd], [2nd]) = (2,3):

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

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

then to avoid ⛔「max ⊢2⊣ = 3」 we need [4th]=4:

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

Note that we fail to match ✅「⟨⋯ ᵃb ⋯⟩, ab=4」. This shows a contradiction.

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

We have verified (2.1). Accordingly, we get

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

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

Now, we consider what [0th] is. By ⛔「Jump(4,5) = 0」, it is not 4. By ⛔「⟨ ⁵ᵗʰ↓ ⁴ᵗʰ↑ ³ʳᵈ↓ ²ⁿᵈ↑ ¹ˢᵗ↓ ⁰ᵗʰ↑ ⟩ after ⟨⇌⟩」, it is not 0. Therefore, it is 1:

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

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

Finally, using ⛔「⟨ ⁵ᵗʰ↓ ⁴ᵗʰ↑ ³ʳᵈ↓ ²ⁿᵈ↑ ¹ˢᵗ↓ ⁰ᵗʰ↑ ⟩ after ⟨⇌⟩」 once more, we finish by

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

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

Q.E.D.

#125034_v2.13