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Proof step 5 from proof mod-neq (previous step)

i * j = i * j0 + j1, i < 0, abs(j1) < abs(i) ⊦ j1 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
i * j = i * j0 + j1, i < 0, abs(j1) + i < 0 ⊦ j1 = 0

Rule argument

Additional information

Used simplifier rules: forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs% ⊦ i < ~ j ↔ i + j < 0; (s)% ⊦ i < 0 → abs(i) = ~ i; (s)