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Proof step 14 from proof div-add-succ (previous step)

(i0 * j) % j = 0 ∧ 0 ≤ j0 - 1 ∧ 0 ≤ i0 * j ∧ 0 < j, j0 < j, 0 < j, 0 ≤ j, 0 ≤ i0, 0 ≤ i0 * j + j0, 0 < i0 * j + j0, 0 < j0, 1 < j ⊦ j0 = 0, (j0 - 1) / j + (i0 * j) / j = i0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: div-mult-03: j ≠ 0 ⊦ (i * j) / j = i;%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%zero-ge: ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j); from specification int-mult%eq: ⊦ i = i + j ↔ j = 0; from specification int-basic2%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1%leone: ⊦ 1 ≤ i ↔ 0 < i; from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%sub-le-01: ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; from specification int-abs%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%zero-left: ⊦ 0 * i = 0; from specification int-mult