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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
abs(j0) < j, 0 < j, i0 < j, j0 < j, i0 * j + j0 < 0, i0 ≤ 0, j0 ≤ 0 ⊦ i0 < 0 ↔ j * (i0 + 1) + j0 ≤ 0

Rule argument

Additional information

Used simplifier rules: 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%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 * j + i = i * (j + 1); (s)