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Proof step 5 from proof modmod-add-01 (previous step)

0 ≤ i, 0 ≤ i0 ⊦ (i % j + i0) % j = (i + i0) % j, 0 = j, 0 < j ∧ 0 ≤ i ∧ 0 ≤ i0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
j < 0, 0 ≤ i, 0 ≤ i0 ⊦ (i % j + i0) % j = (i + i0) % j

Rule argument

Additional information

Used simplifier rules: ⊦ i < 0 ∧ 0 ≤ j → i < j; (s)%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% ⊦ i < j → ¬ j < i; (s)% ⊦ ¬ i < j → (i = j ↔ ¬ j < i); (ss)% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; (s)