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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
j0 * j + j = i0 + 1, 0 < j, 1 < j, 0 ≤ j0 ⊦ j = 0, i0 / j = j0

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%zero-ge: ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j); from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult