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Proof step 2 from proof zero-le-two (previous step)

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

Lemma zero-le-case