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Proof step 5 from proof div-add-both (previous step)

i % j = 0, ∃ j0. i0 = j0 * j ⊦ (i + i0) / j = i / j + i0 / j, j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
i % j = 0 ⊦ (i + j0 * j) / j = i / j + j0, j = 0

Rule argument

Additional information

Used simplifier rules: div-mult-03: j ≠ 0 ⊦ (i * j) / j = i;%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult