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Proof step 10 from proof square-ex (previous step)

(j + 1) * (j + 1) = i0 + 1, 0 < i0, 0 < j, j * j ≤ 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 11:
(j + 1) * (j + 1) = i0 + 1, 0 < i0, 0 < j, j * j ≤ i0, 0 ≤ i0, (2 + j) * (2 + j) ≤ i0 + 1 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i ≤ i; (s)% ⊦ i + j ≤ i0 + j ↔ i ≤ i0; (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ 0 < i ∧ 0 < j → 0 < i + j; (s)