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Proof step 1 from proof mod-zero-06 (conclusion)

⊦ 0 ≤ i ∧ j ≠ 0 → ¬ i % j < 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: mod-ls-same-01: j ≠ 0 ⊦ i % j < i ↔ 0 < i ∧ (abs(j) = 1 ∨ abs(j) ≤ i);%succ-01: ⊦ i +1 = i + 1; from specification int-basic2%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%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%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs