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Proof step 4 from proof mult-02 (previous step)

i = i1 * j0 + j1 ∧ 0 ≤ j1 ∧ j1 < j0, 0 ≤ i0, i0 < j, 0 ≤ i, 0 < j0, 0 < j ⊦ (i * j + i0) / (j0 * j) = i1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
0 < j0, i0 < j, 0 ≤ i0, j1 < j0, 0 < j, 0 ≤ i1, 0 ≤ j1 ⊦ ((i1 * j0 + j1) * j + i0) / (j0 * j) = i1

Rule argument

Additional information

Used simplifier rules: mone-06: ⊦ ~1 < i ↔ 0 ≤ i; from specification int-abs%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%numlef-01: ⊦ (* i) ≤ j → i - 1 < j; (forward) 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%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%zero-ge-add: ⊦ 0 ≤ j1 ∧ j1 < j → (0 ≤ i0 * j + j1 ↔ 0 ≤ i0); from specification int-mult