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Proof step 1 from proof zero-ge-03 (conclusion)

⊦ i ≤ 0 → (0 ≤ i * j ↔ ¬ ¬(i = 0 ∨ j ≤ 0))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
i ≤ 0 ⊦ 0 ≤ i * j ↔ i = 0 ∨ j ≤ 0

Rule argument

Additional information

Used simplifier rules: forward-le-04: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward) from specification int-abs