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Proof step 9 from proof div-ls-mone (previous step)

¬ j0 < 0, i0 * j + j0 < j, i0 * j + j0 < 0, i0 < j, i0 ≤ 0, j0 < j, j0 ≤ 0, 0 < j, abs(j0) < j, (i0 * j + j0) + 1 < j ⊦ i0 < ~1 ↔ (i0 * j + j0) + 2 * j ≤ 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: succ-01: ⊦ i +1 = i + 1; from specification int-basic2%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%lsone: ⊦ i < 1 ↔ i ≤ 0; from specification int-abs%ls-same: ⊦ 0 < i → (i * j < i ↔ j < 1); from specification int-mult%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%zero-gels-03: ⊦ i < 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%zero-ls: ⊦ 0 < i → (i * j < 0 ↔ j < 0); 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%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1%anti: ⊦ i ≤ j → (j ≤ i ↔ i = j); from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%Abs-pos: ⊦ ¬ j < 0 → abs(j) = j; from specification int-abs