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Proof step 2 from proof div-elim1 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
j < 0, 0 ≤ i ⊦ i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 + j < 0

Rule argument

Additional information

Used simplifier rules: ⊦ i < 0 ∧ 0 ≤ j → i < j; (s)%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (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% ⊦ i < ~ j ↔ i + j < 0; (s)% ⊦ i < 0 → abs(i) = ~ i; (s)