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Proof step 6 from proof div-add-nosucc (previous step)

i + j = i0 + 1, 0 < j, 1 < j, 0 ≤ i, ∃ j0. i = j0 * j ⊦ i0 / j = i / j, j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
j0 * j + j = i0 + 1, 1 < j, 0 < j, 0 ≤ j0 ⊦ i0 / j = j0, j = 0

Rule argument

Additional information

Used simplifier rules: div-mult-03: j ≠ 0 ⊦ (i * j) / j = i;%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs% ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j); (s)%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult