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Proof step 2 from proof div-add1-posneg-pos (previous step)

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

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: div-add1-negpos-pos: ⊦ i % j = 0 ∧ 0 ≤ i + i0 ∧ i < 0 ∧ 0 ≤ i0 ∧ 0 < j → (i + i0) / j = i / j + i0 / j;% ⊦ i + j = i + i0 ↔ j = i0; (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2