Rearrange the digits in ⟨125034⟩ to meet the rules below.
⟨ ⁵ᵗʰ▨ ⁴ᵗʰ▨ ³ʳᵈ▨ ²ⁿᵈ▨ ¹ˢᵗ▨ ⁰ᵗʰ▨ ⟩
✅Match
⟨ ▧ ▢ ▢ − − − ⟩, ▧ ≥ Σ▢
{p3, p2, p1} = ? + {0,1,2}
Sim⟨ ⁵ᵗʰ5 ⁴ᵗʰ4 ³ʳᵈ1 ²ⁿᵈ0 ¹ˢᵗ2 ⁰ᵗʰ3 ⟩ ≤ 1
median {p5, p4, p3} = 1
#125034_v2.13
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│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ 0 │ │ │ │ │▒
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Step 2 │ │ 0 │ 1 │ │ │ │▒
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Step 3 │ │ 0 │ 1 │ 2 │ │ │▒
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Step 4 │ │ 0 │ 1 │ 2 │ 3 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 4 │ 0 │ 1 │ 2 │ 3 │ │▒
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Step 6 │ 4 │ 0 │ 1 │ 2 │ 3 │ 5 │▒
└───┴───┴───┴───┴───┴───┘▒
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Proof of 2026-05-05 WR
══════════════════════
Notation: if Nth -> a, then we write pN = a.
✅「median {p5, p4, p3} = 1」 implies
(1) both 0 and 1 are in {p5, p4, p3}.
A fortiori, 0 = p5 | p4 | p3.
(2) We claim that indeed 0 = p4.
------------------------------
For, if 0 = p5, then we cannot match ✅「⟨ ▧ ▢ ▢ − − − ⟩, ▧ ≥ Σ▢」. Else if 0 = p3:
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│5th│4th│ 3▲│2nd│1st│0th│
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│ │ │ 0 │ │ │ │
└───┴───┴───┴───┴───┴───┘
then ✅「{p3, p2, p1} = ? + {0,1,2}」 implies {p3, p2, p1} = {0,1,2}:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ 0 │1 2│1 2│ │
└───┴───┴───┴───┴───┴───┘
We cannot match this and (1) at the same time, which is a contradiction.
------------------------------
We have verified (2) and get
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│5th│ 4■│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ 0 │ │ │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
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--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 5 │ │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
In view of (1), we have 1 = p5 or p3:
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│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ ▬ │ 0 │ ▬ │ │ │ │
└───┴───┴───┴───┴───┴───┘
If it is p5, then we cannot match ✅「⟨ ▧ ▢ ▢ − − − ⟩, ▧ ≥ Σ▢」. Therefore, it is p3.
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3■│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 0 │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ 0 │ 1 │ │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
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--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ 5 │ │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
Now, to match ✅「{p3, p2, p1} = ? + {0,1,2}」, we need
(3) {p2, p1} = {2,3}.
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│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 0 │ 1 │2 3│2 3│ │
└───┴───┴───┴───┴───┴───┘
In view of ✅「Sim⟨ ⁵ᵗʰ5 ⁴ᵗʰ4 ³ʳᵈ1 ²ⁿᵈ0 ¹ˢᵗ2 ⁰ᵗʰ3 ⟩ ≤ 1」, as we already have an agreed positional digit 1=p3, to match this pattern we cannot have 2=p1. Therefore, we have
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│5th│4th│3rd│ 2■│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 0 │ 1 │ │ │ │▒
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Step 3 │ │ 0 │ 1 │ 2 │ │ │▒
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Step 4 │ │ 0 │ 1 │ 2 │ 3 │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ 5 │ │ │ 4 │
└───┴───┴───┴───┴───┴───┘
Using ✅「Sim⟨ ⁵ᵗʰ5 ⁴ᵗʰ4 ³ʳᵈ1 ²ⁿᵈ0 ¹ˢᵗ2 ⁰ᵗʰ3 ⟩ ≤ 1」 once more, we finish by
┌───┬───┬───┬───┬───┬───┐
│ 5■│4th│3rd│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 0 │ 1 │ 2 │ 3 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 4 │ 0 │ 1 │ 2 │ 3 │ │▒
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Step 6 │ 4 │ 0 │ 1 │ 2 │ 3 │ 5 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.13