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Theorem square-inject

⊦ 0 ≤ i ∧ 0 ≤ j ∧ 0 ≤ i0 ∧ 0 ≤ j0 → (((i + j) * (i + j + 1)) / 2 + j = ((i0 + j0) * (i0 + j0 + 1)) / 2 + j0 ↔ i = i0 ∧ j = j0)

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