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Proof step 1 from proof mod-nls-02 (conclusion)

⊦ i0 ≤ i ∧ i0 < 0 ∧ j ≠ 0 → ¬ i % j < i0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
i0 < i, i % j < 0, i % j < ~1, i0 < 0, i % j < i0, i0 ≤ i, 0 < i, i0 + 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);%lst-04: ⊦ i + j < i0 ∧ j0 < j → i + j0 < i0; from specification int-basic2%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%numls-03: ⊦ i + (* j) < (* j0) ↔ i < j0 - j; from specification int-abs%succ-01: ⊦ i +1 = i + 1; from specification int-basic2%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (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%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs