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Proof step 11 from proof div-est-num (previous step)

i = i0 * j + j2, i1 * j + j1 < 0, i1 < i0, j1 < j, i1 < j, i < 0, 0 < j, i1 < j0, abs(j1) < j, abs(j2) < j, j1 ≤ 0, i1 ≤ 0, j0 ≤ i0, i ≤ i1 * j + j1, j2 ≤ 0, i0 ≤ 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)%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 < 0 ∧ j0 < j ∧ 0 < j → i * j + j0 < 0; (s)%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%forward-le-04: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward) from specification int-abs%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