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Proof step 14 from proof square-ex (previous step)

0 ≤ i0, 0 < i0 → 0 < j ∧ j * j ≤ i0 ∧ i0 < (j + 1) * (j + 1) ⊦ 0 < i0, ∃ j0. 0 < j0 ∧ j0 * j0 ≤ i0 + 1 ∧ i0 + 1 < (j0 + 1) * (j0 + 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%anti: ⊦ i ≤ j → (j ≤ i ↔ i = j); from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs