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Proof step 1 from proof mod-add1-negpos-neg (conclusion)

⊦ i % j = 0 ∧ i + i0 < 0 ∧ i < 0 ∧ 0 ≤ i0 ∧ i0 % j ≠ 0 ∧ 0 < j → (i + i0) % j = i0 % j - j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
i % j = 0, 0 < j, 0 ≤ i0, i < 0, i + i0 < 0 ⊦ i0 % j = (i + i0) % j + j, i0 % j = 0

Rule argument

Additional information

Used simplifier rules: add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%ls: ⊦ i + j < j ↔ i < 0; from specification int-basic2%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; 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%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs