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Proof step 2 from proof mod-add-01 (previous step)

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 3:
0 < i, j < 0, i + j ≤ 0 ⊦ (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% ⊦ i < 0 ∧ 0 ≤ j → i < j; (s)% ⊦ i + j < j ↔ i < 0; (s)%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (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% ⊦ i ≤ ~ j ↔ i + j ≤ 0; (s)% ⊦ i < 0 → abs(i) = ~ i; (s)