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Proof step 7 from proof div-est-num (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
i1 * j + j1 < j, i1 * j + j1 < 0, i1 < i / j, i1 < j, i1 ≤ 0, j1 < j, j1 ≤ 0, i + 1 < j, i < j, i < 0, 0 < j, j0 ≤ i / j, i ≤ i1 * j + j1, i1 < j0, (i1 * j + j1) + 1 < j, abs(j1) < j ⊦

Rule argument

Additional information

Used simplifier rules: lst-03: ⊦ i + j < i0 ∧ ¬ j < j0 → i + j0 < i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%succ-01: ⊦ i +1 = i + 1; from specification int-basic2%forward-le-04: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward) from specification int-abs%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%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