Rearrange the digits in ⟨125034⟩ to meet the rules below.
⟨5th 4th 3rd 2nd 1st 0th⟩
✅Match
⟨ ⁴ᵗʰb ²ⁿᵈc ⁰ᵗʰa ⟩, a > b > c
⟨⋯ 5 ⋯ 2 ⋯⟩
3rd → a, 0th → b, |a-b|=2
⛔Avoid
min {p5, p3, p0} = 2
⟦4,5⟧ ∋ 0,2
#125034_v2.10
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ │ │ 3 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ 1 │ │ │ │ 3 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ 1 │ │ 0 │ │ 3 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ │ 1 │ 5 │ 0 │ │ 3 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ │ 1 │ 5 │ 0 │ 2 │ 3 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 4 │ 1 │ 5 │ 0 │ 2 │ 3 │▒
└───┴───┴───┴───┴───┴───┘▒
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Proof of 2025-09-30 WR
══════════════════════
We consider what [0th] is. By ✅「⟨ ⁴ᵗʰb ²ⁿᵈc ⁰ᵗʰa ⟩, a > b > c」, we have
(1) [0th] = 5|4|3|2.
(1.1) We show that [0th] = 3 actually.
------------------------------
(2.1) If [0th] = 5:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ │ │ │ 5 │
└───┴───┴───┴───┴───┴───┘
then we would fail to match ✅「⟨⋯ 5 ⋯ 2 ⋯⟩」, which is a contradiction.
(2.2) Else if [0th] = 4, then it follows from ✅「3rd → a, 0th → b, |a-b|=2」 that [3rd] = 2:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3▲│2nd│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ 2 │ │ │ 4 │
└───┴───┴───┴───┴───┴───┘
Then, to match ✅「⟨⋯ 5 ⋯ 2 ⋯⟩」, we have to place 5 at 5th or 4th. But both give contradictions: if 5 = [5th], then we fail to avoid ⛔「min {p5, p3, p0} = 2」; else if 5 = [4th], then we fail to match ✅「⟨ ⁴ᵗʰb ²ⁿᵈc ⁰ᵗʰa ⟩, a > b > c」.
(2.3) Lastly, if [0th] = 2, then in view of ✅「⟨ ⁴ᵗʰb ²ⁿᵈc ⁰ᵗʰa ⟩, a > b > c」, we have
┌───┬───┬───┬───┬───┬───┐
│5th│ 4▲│3rd│ 2▲│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 1 │ │ 0 │ │ 2 │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ 5 │ │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
Then, to match ✅「3rd → a, 0th → b, |a-b|=2」, we need [3rd] = 4:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3▲│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 1 │ 4 │ 0 │ │ 2 │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ 5 │ │ 3 │ │
└───┴───┴───┴───┴───┴───┘
Observe that we would fail to avoid ⛔「min {p5, p3, p0} = 2」, which is a contradiction.
------------------------------
We have verified (1.1). Accordingly, we get
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ │ │ 3 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 5 │ 0 │ │ 4 │
└───┴───┴───┴───┴───┴───┘
Next, we consider what [4th] is. By ✅「⟨ ⁴ᵗʰb ²ⁿᵈc ⁰ᵗʰa ⟩, a > b > c」, we have
(3) [4th] = 2|1.
(3.1) We claim that [4th] = 1.
------------------------------
For, suppose on the contrary [4th] = 2, then by ✅「⟨⋯ 5 ⋯ 2 ⋯⟩」 we have [5th] = 5 as well:
┌───┬───┬───┬───┬───┬───┐
│ 5▲│ 4▲│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 5 │ 2 │ │ │ │ 3 │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ │ 0 │ │ 4 │
└───┴───┴───┴───┴───┴───┘
Then, to match ✅「3rd → a, 0th → b, |a-b|=2」, we need [3rd] = 1:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3▲│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 5 │ 2 │ 1 │ │ │ 3 │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ 0 │ │ 4 │
└───┴───┴───┴───┴───┴───┘
So, using ✅「⟨ ⁴ᵗʰb ²ⁿᵈc ⁰ᵗʰa ⟩, a > b > c」, we reach
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│ 2▲│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 5 │ 2 │ 1 │ 0 │ 4 │ 3 │
└───┴───┴───┴───┴───┴───┘
It however matches ⛔「⟦4,5⟧ ∋ 0,2」.
------------------------------
We have verified (3.1). As a result, we get
┌───┬───┬───┬───┬───┬───┐
│5th│ 4■│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ │ │ │ 3 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ 1 │ │ │ │ 3 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ 5 │ 0 │ │ 4 │
└───┴───┴───┴───┴───┴───┘
Then, it follows from ✅「⟨ ⁴ᵗʰb ²ⁿᵈc ⁰ᵗʰa ⟩, a > b > c」 that [2nd] = 0:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│ 2■│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 1 │ │ │ │ 3 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ 1 │ │ 0 │ │ 3 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ 5 │ │ │ 4 │
└───┴───┴───┴───┴───┴───┘
On the other hand, using ✅「3rd → a, 0th → b, |a-b|=2」, we get [3rd] = 5 :
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3■│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 1 │ │ 0 │ │ 3 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ │ 1 │ 5 │ 0 │ │ 3 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ │ │ │ 4 │
└───┴───┴───┴───┴───┴───┘
Finally, using ✅「⟨⋯ 5 ⋯ 2 ⋯⟩」, we finish by
┌───┬───┬───┬───┬───┬───┐
│ 5■│4th│3rd│2nd│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 1 │ 5 │ 0 │ │ 3 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ │ 1 │ 5 │ 0 │ 2 │ 3 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 4 │ 1 │ 5 │ 0 │ 2 │ 3 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.10