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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

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