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Proof step 6 from proof div-ls-zero (previous step)

i = i0 * j + j0 ∧ j0 ≤ 0 ∧ abs(j0) < j ∧ i0 ≤ 0, i < 0, 0 < j ⊦ i0 < 0 ↔ i + j ≤ 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
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 < 0 ↔ (i0 * j + j0) + j ≤ 0

Rule argument

Additional information

Used simplifier rules: 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-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