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

⊦ 0 ≤ i * j ↔ i = 0 ∨ j = 0 ∨ 0 ≤ i ∧ 0 ≤ j ∨ i ≤ 0 ∧ j ≤ 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs