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Proof step 17 from proof exdivmod (previous step)

j < 0, i ≤ 0 → i + j1 = j * i1 ∧ 0 ≤ i1 ∧ 0 ≤ j1 ∧ j1 + j < 0 ⊦ ¬ 0 < i, j = 0, ∃ i0, j0. i = i0 * j + j0 ∧ abs(j0) + j < 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
j < 0, 0 < i, i ≤ 0 → i + j1 = j * i1 ∧ 0 ≤ i1 ∧ 0 ≤ j1 ∧ j1 + j < 0 ⊦ j = 0, ∃ i0, j0. i = i0 * j + j0 ∧ abs(j0) + j < 0

Rule argument

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs