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Proof step 4 from proof mod-grz-01 (previous step)

¬(0 < j ∧ i < 0), i < 0, 0 < i % j, i + 1 < i % j, abs(j) = 1 ∨ abs(j) + i ≤ 0 ⊦ j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
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-pos-not-neg: ⊦ 0 < i1 ∧ 0 < i % i1 → ¬ i < 0;