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Proof step 4 from proof modmod-add-04 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
0 < j, 0 ≤ i0, 0 ≤ i ⊦ (i % j + i0 % j) % j = (i + i0) % j

Rule argument

Additional information

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