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Proof step 23 from proof exdivmod (previous step)

0 ≤ i ∧ 0 < j → i = i0 * j + j0 ∧ 0 ≤ i0 ∧ 0 ≤ j0 ∧ j0 < j ⊦ j < 0, 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

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%abs-16: ⊦ 0 < j → abs(j) = j; from specification int-abs%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1%lst-01: ⊦ ¬ i < j → (i ≠ j ↔ j < i); from specification int-basic1