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Proof step 2 from proof zero-ge-add-02 (previous step)

0 < j0, 0 ≤ i + j ⊦ 0 ≤ i * j0 + j * j0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
i * j0 + j * j0 < j0, i * j0 + j * j0 < i + j, i * j0 + j * j0 < 0, 0 < j0, (i * j0 + j * j0) + 1 < j0, 0 ≤ i + j ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ i +1 = i + 1; (s)%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1% ⊦ i ≤ j ↔ ¬ j < i; (ss)