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Proof step 2 from proof abs-ls (previous step)

j < 0, i < 0, j < j0, j0 < 0 ⊦ abs(j0) < i * j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: ⊦ i +1 = i + 1; (s)%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)% ⊦ ~ i < j ↔ 0 < j + i; (s)% ⊦ 0 - i = ~ i; (s)% ⊦ j < 0 → abs(j) = 0 - j; (s)