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Proof step 3 from proof square-interval (previous step)

i * (i + 1) ≤ j * (j + 1) + 2 * j0, j * (j + 1) + 2 * j0 < (i + 1) * (i + 2), j0 ≤ j, 0 ≤ i, 0 ≤ j0 ⊦ i = j

A spec-lemma