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

(i1 * j) % j = 0, j0 < j, 0 < j, 0 ≤ i1, 0 ≤ j0, 0 ≤ i0, 0 ≤ i1 * j, 0 ≤ j0 + i0, (j0 + i0 % j) / j = 1 ∨ (j0 + i0 % j) / j = 0 ⊦ (i1 + i0 / j) + (j0 + i0 % j) / j = (i1 * j) / j + (j0 + i0) / j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
j0 < j, 0 < j, 0 ≤ i1, 0 ≤ j0, 0 ≤ i0, 0 ≤ j0 + i0, (j0 + i0 % j) / j = 1 ∨ (j0 + i0 % j) / j = 0 ⊦ i0 / j + (j0 + i0 % j) / j = (j0 + i0) / j

Rule argument

Additional information

Used simplifier rules: mod-mult-04: j ≠ 0 ⊦ (i * j) % j = 0;%div-mult-03: j ≠ 0 ⊦ (i * j) / j = i;%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; (s)% ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j); (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i + j = i + i0 ↔ j = i0; (s)%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult% ⊦ i * 0 = 0; (s)