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Proof step 8 from proof div-add-neg (previous step)

i0 = i2 * j + j1, j1 = 0, (i1 * j + j0) + 1 < j, abs(j0) < j, i0 < j, i0 < 0, 0 < j, i1 * j + j0 < 0, i0 + 1 < j, j0 < j, i1 * j + j0 < j, abs(j1) < j, j0 ≤ 0, j1 ≤ 0 ⊦ ((i1 * j + j0) + i0) / j = i1 + i2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
i2 < j, i2 < 0, i1 * j + j0 < 0, 1 + i2 * j < j, j0 < j, i1 * j + j0 < j, 0 < j, (i1 * j + j0) + 1 < j, abs(j0) < j, i2 + 1 < j, i2 ≤ 0, j0 ≤ 0 ⊦ (j * (i2 + i1) + j0) / j = i1 + i2

Rule argument

Additional information

Used simplifier rules: ⊦ i +1 = i + 1; (s)%trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1% ⊦ i < 1 ↔ i ≤ 0; (s)% ⊦ 0 < i → (i * j < i ↔ j < 1); (s)% ⊦ 0 < i → (i * j < 0 ↔ j < 0); (s)%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i + 0 = i; (s)% ⊦ i * j + i * j0 = i * (j + j0); (s)