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Proof step 9 from proof square-ex (previous step)

(j + 1) * (j + 1) = i0 + 1, 0 < j, j * j ≤ i0, 0 ≤ i0, 0 < i0 ⊦ 0 < j + 1 ∧ (j + 1) * (j + 1) ≤ i0 + 1 ∧ i0 + 1 < ((j + 1) + 1) * ((j + 1) + 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
(j + 1) * (j + 1) = i0 + 1, 0 < i0, 0 ≤ i0, j * j ≤ i0, 0 < j, (2 + j) * (2 + j) ≤ i0 + 1 ⊦

Rule argument

Additional information

Used simplifier rules: lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%ler: ⊦ i ≤ i; from specification int-abs%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%numle-01: ⊦ i + (* j) ≤ i0 + (* j0) ↔ i ≤ i0 + (j0 - j); from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%zero-02: ⊦ 0 < i ∧ 0 < j → 0 < i + j; from specification int-basic2