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Proof step 1 from proof mod-grz-01 (conclusion)

⊦ i < 0 ∧ j ≠ 0 → ¬ 0 < i % j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
i < 0, 0 < i % j, i + 1 < i % j, abs(j) = 1 ∨ abs(j) + i ≤ 0 ⊦ j = 0

Rule argument

Additional information

Used simplifier rules: mod-ls-same-02: j ≠ 0 ⊦ i < i % j ↔ i < 0 ∧ (abs(j) = 1 ∨ abs(j) ≤ abs(i));%negle-07: ⊦ i ≤ ~ j ↔ i + j ≤ 0; from specification int-abs%abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%succ-01: ⊦ i +1 = i + 1; from specification int-basic2%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1