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Proof step 14 from proof div-add1-negpos-neg-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ i < 0 ∧ 0 ≤ j → i < j; (s)%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%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs% ⊦ ~ i < j ↔ 0 < j + i; (s)% ⊦ i < 0 → abs(i) = ~ i; (s)% ⊦ i < 0 ∧ 0 ≤ j → i ≤ j; (s)% ⊦ i + j < j ↔ i < 0; (s)%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i < j → ¬ j < i; (s)% ⊦ i ≤ j ↔ ¬ j < i; (ss)%commutativity: ⊦ i * j = j * i; from specification int-mult% ⊦ i - j < i0 ↔ i < j + i0; (s)% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; (s)% ⊦ i - j = i0 ↔ i = i0 + j; (s)% ⊦ i + 0 = i; (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 - i0) = i + j - i0; (s)