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Proof step 2 from proof mod-nls-02 (previous step)

i % j < i0, i0 < 0, i0 ≤ i ⊦ j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
i % j < i0, i0 < 0, 0 < i, i0 ≤ 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);%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