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Proof step 9 from proof div-zero-02 (previous step)

abs(i) < abs(j) ⊦ i / j = 0, 0 < j ∧ 0 ≤ i ∧ i < j, 0 < j ∧ i < 0 ∧ abs(i) < j, 0 < ~ j ∧ 0 ≤ i ∧ i < ~ j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
abs(i) < abs(j) ⊦ i / j = 0, 0 < j ∧ 0 ≤ i ∧ i < j, j < 0 ∧ 0 ≤ i ∧ i + j < 0, 0 < j ∧ i < 0 ∧ ~ i < j

Rule argument

Additional information

Used simplifier rules: ⊦ i < 0 → abs(i) = ~ i; (s)% ⊦ i < ~ j ↔ i + j < 0; (s)% ⊦ 0 < ~ i ↔ i < 0; (s)