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Proof step 1 from proof div-div (conclusion)

⊦ i | j ∧ j | i ∧ i ≤ 0 ∧ j ≤ 0 → (i ≠ j ↔ false)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
i | j, j | i, i ≤ 0, j ≤ 0 ⊦ i = j

Rule argument

Additional information

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