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Proof step 24 from proof exdivmod (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
i < 0, 0 < j ⊦ 0 ≤ i, ∃ i1, j1. i = i1 * j + j1 ∧ abs(j1) < j

Rule argument

Additional information

Used simplifier rules: forward-le-05: ⊦ 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% ⊦ 0 < j → abs(j) = j; (s)% ⊦ i < j → ¬ j < i; (s)% ⊦ ¬ i < j → (i = j ↔ ¬ j < i); (ss)