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Proof step 8 from proof div-add1-negpos-neg-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: ⊦ i < 0 ∧ 0 ≤ j → i < j; (s)% ⊦ i + j < j ↔ i < 0; (s)% ⊦ 0 < i → (i * j < i * j0 ↔ j < j0); (s)% ⊦ i + j < i + i0 ↔ j < i0; (s)%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs% ⊦ i ≠ j → (i ≤ j ↔ i < j); (s)%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% ⊦ i * j + i * j0 = i * (j + j0); (s)