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Proof step 1 from proof mod-add-01 (conclusion)

⊦ 0 < i ∧ j < 0 ∧ i ≤ abs(j) → (i + j) % i = j % i

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

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%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%ls: ⊦ i + j < j ↔ i < 0; from specification int-basic2%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-le-04: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward) from specification int-abs%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%negle-07: ⊦ i ≤ ~ j ↔ i + j ≤ 0; from specification int-abs%abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs