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Proof step 1 from proof div-add-pos (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: mod-less-01: ⊦ 0 < j → i % j < j;%lf-03: ⊦ i < j ∧ i = i0 → i0 < j; (forward) from specification int-basic1