Rearrange the digits in ⟨125034⟩ to meet the rules below.
⟨5th 4th 3rd 2nd 1st 0th⟩
✅Match
1st → a, 0th → b, |a-b|=1
⟨ ⁴ᵗʰa ³ʳᵈb ⁰ᵗʰc ⟩, (abc)₁₀ ≥ 153
⟦0,2⟧ ∋ 3,4
⛔Avoid
⟨ ⁵ᵗʰ↑ ⁴ᵗʰ↓ ⟩ after 3×⟨→⟩
{p4, p2, p0} = ? + {0,1,2}
2nd → 0|4
#125034_v2.9
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│▒
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Step 1 │ │ │ 4 │ │ │ │▒
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Step 2 │ 5 │ │ 4 │ │ │ │▒
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Step 3 │ 5 │ 2 │ 4 │ │ │ │▒
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Step 4 │ 5 │ 2 │ 4 │ 3 │ │ │▒
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Step 5 │ 5 │ 2 │ 4 │ 3 │ 1 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 5 │ 2 │ 4 │ 3 │ 1 │ 0 │▒
└───┴───┴───┴───┴───┴───┘▒
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Proof of 2025-08-05 WR
══════════════════════
Notation: if nth -> a, then we write [nth] = a.
Firstly, we consider where to place 4. By ✅「⟦0,2⟧ ∋ 3,4」, it is not in corners (5th or 0th). By ⛔「2nd → 0|4」, it is not at 2nd.
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
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│ / │ │ │ / │ │ / │
└───┴───┴───┴───┴───┴───┘
If 4 = [1st]:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│ 1▲│0th│
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│ / │ │ │ / │ 4 │ / │
└───┴───┴───┴───┴───┴───┘
then to match ✅「1st → a, 0th → b, |a-b|=1」, we need [0th] = 3|5. However, in both cases, we would fail to match ✅「⟦0,2⟧ ∋ 3,4」. Therefore, 4 != [1st].
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ / │ │ │ / │ / │ / │
└───┴───┴───┴───┴───┴───┘
(1) It follows that 4 = [4th] | [3rd], and we claim that indeed 4 = [3rd].
For, if on the contrary 4 = [4th]:
┌───┬───┬───┬───┬───┬───┐
│5th│ 4▲│3rd│2nd│1st│0th│
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│ │ 4 │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
then, to match ✅「⟦0,2⟧ ∋ 3,4」, we need [5th] = 0|2:
┌───┬───┬───┬───┬───┬───┐
│ 5▲│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
(2.1) │ 0 │ 4 │ │ │ │ │
├───┼───┼───┼───┼───┼───┤
(2.2) │ 2 │ 4 │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
(3) We show that both cases lead to contradictions.
------------------------------
(3.1) If case (2.1) holds, then to avoid ⛔「⟨ ⁵ᵗʰ↑ ⁴ᵗʰ↓ ⟩ after 3×⟨→⟩」, we need [1st] = 5:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 0 │ 4 │ │ │ 5 │ │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
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│ 1 │ 2 │ │ │ 3 │ │
└───┴───┴───┴───┴───┴───┘
But then we would fail to match ✅「1st → a, 0th → b, |a-b|=1」.
(3.2) Else if case (2.2) holds:
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│ 5▲│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ 4 │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ 5 │ 0 │ 3 │ │
└───┴───┴───┴───┴───┴───┘
then to match ✅「1st → a, 0th → b, |a-b|=1」, we have {[1st], [0th]} = {0,1}, whence {[3th], [2nd]} = {3,5}:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ 4 │3 5│3 5│0 1│0 1│
└───┴───┴───┴───┴───┴───┘
We would match ⛔「⟨ ⁵ᵗʰ↑ ⁴ᵗʰ↓ ⟩ after 3×⟨→⟩」, which is a contradiction.
------------------------------
We have verified (3) and (1). Accordingly, we get 4 = [3rd]:
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│5th│4th│ 3■│2nd│1st│0th│▒
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Step 1 │ │ │ 4 │ │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
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--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 5 │ 0 │ 3 │ │
└───┴───┴───┴───┴───┴───┘
Next, we consider where to place 5. To match ✅「1st → a, 0th → b, |a-b|=1」, it cannot be at 1st or 0th.
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ 4 │ │ / │ / │
└───┴───┴───┴───┴───┴───┘
(4) So, 5 = [5th] | [4th] | [2nd]. We proceed to show that 5 != [2nd].
------------------------------
Suppose on the contrary 5 = [2nd]:
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│5th│4th│3rd│ 2▲│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
(4.1) │ │ │ 4 │ 5 │ │ │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ │ 0 │ 3 │ │
└───┴───┴───┴───┴───┴───┘
Then, let us consider where to place 0:
• If 0 is at 1st or 0th in (4.1), then to match ✅「1st → a, 0th → b, |a-b|=1」, we need {0,1} = {[1st], [0th]}:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ 4 │ 5 │0 1│0 1│
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ │ │ 3 │ │
└───┴───┴───┴───┴───┴───┘
But then we would match ⛔「⟨ ⁵ᵗʰ↑ ⁴ᵗʰ↓ ⟩ after 3×⟨→⟩」, which is a contradiction.
• Else if 0 = [4th] in (4.1), then we cannot match ✅「⟨ ⁴ᵗʰa ³ʳᵈb ⁰ᵗʰc ⟩, (abc)₁₀ ≥ 153」.
It follows that 0 = [5th] in (4.1):
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│ 5▲│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 0 │ │ 4 │ 5 │ │ │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ │ │ 3 │ │
└───┴───┴───┴───┴───┴───┘
We consider what [4th] is. In view of ✅「⟨ ⁴ᵗʰa ³ʳᵈb ⁰ᵗʰc ⟩, (abc)₁₀ ≥ 153」, it is not 1; and to match ✅「⟦0,2⟧ ∋ 3,4」, it is not 2. Therefore [4th] = 3:
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│5th│ 4▲│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 0 │ 3 │ 4 │ 5 │ │ │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Note that we would match ⛔「⟨ ⁵ᵗʰ↑ ⁴ᵗʰ↓ ⟩ after 3×⟨→⟩」, which is a contradiction.
------------------------------
We have verified (4). Accordingly, we have 4 = [5th] | [4th]:
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│5th│4th│3rd│2nd│1st│0th│
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(5) │ │ 5 │ 4 │ │ │ │
├───┼───┼───┼───┼───┼───┤
(6) │ 5 │ │ 4 │ │ │ │
└───┴───┴───┴───┴───┴───┘
Suppose case (5) holds. Then, to match ✅「⟦0,2⟧ ∋ 3,4」 and avoid ⛔「⟨ ⁵ᵗʰ↑ ⁴ᵗʰ↓ ⟩ after 3×⟨→⟩」, we have
┌───┬───┬───┬───┬───┬───┐
│ 5▲│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ 5 │ 4 │ │ │ │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ │ 0 │ 3 │ │
└───┴───┴───┴───┴───┴───┘
To match ✅「1st → a, 0th → b, |a-b|=1」, we then have
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│ 2▲│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ 5 │ 4 │ 3 │0 1│0 1│
└───┴───┴───┴───┴───┴───┘
However, it would match ⛔「⟨ ⁵ᵗʰ↑ ⁴ᵗʰ↓ ⟩ after 3×⟨→⟩」, which is a contradiction.
It shows that case (5) does not hold, and case (6) holds indeed:
┌───┬───┬───┬───┬───┬───┐
│ 5■│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ 4 │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ 5 │ │ 4 │ │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ │ 0 │ 3 │ │
└───┴───┴───┴───┴───┴───┘
Given that, to match ✅「⟦0,2⟧ ∋ 3,4」, we need [4th] = 0|2. In view of ✅「⟨ ⁴ᵗʰa ³ʳᵈb ⁰ᵗʰc ⟩, (abc)₁₀ ≥ 153」, we get:
┌───┬───┬───┬───┬───┬───┐
│5th│ 4■│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 5 │ │ 4 │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 5 │ 2 │ 4 │ │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ │ 0 │ 3 │ │
└───┴───┴───┴───┴───┴───┘
To match ✅「1st → a, 0th → b, |a-b|=1」, we need to use 0,1. It follows that [3rd] = 3:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│ 2■│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 5 │ 2 │ 4 │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 5 │ 2 │ 4 │ 3 │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ │ 0 │ │ │
└───┴───┴───┴───┴───┴───┘
Finally, to avoid ⛔「{p4, p2, p0} = ? + {0,1,2}」, we finish by
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│ 1■│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 5 │ 2 │ 4 │ 3 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 5 │ 2 │ 4 │ 3 │ 1 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 5 │ 2 │ 4 │ 3 │ 1 │ 0 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.9