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Proof step 10 from proof exdivmod (previous step)

j < 0, 0 ≤ i ∧ 0 < j → i = i0 * j + j0 ∧ 0 ≤ i0 ∧ 0 ≤ j0 ∧ j0 < j ⊦ j = 0, ¬ i < 0 ∧ ¬ j < 0, ∃ i0, j0. i = i0 * j + j0 ∧ abs(j0) < abs(j)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
j < 0, 0 ≤ i ∧ 0 < j → i = i0 * j + j0 ∧ 0 ≤ i0 ∧ 0 ≤ j0 ∧ j0 < j ⊦ j = 0, ∃ i1, j1. i = i1 * j + j1 ∧ abs(j1) < ~ j

Rule argument

Additional information

Used simplifier rules: forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs