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Proof step 6 from proof exdivmod (previous step)

¬ i < 0 ∧ ¬ j < 0, 0 ≤ i ∧ 0 < j → i = i0 * j + j0 ∧ 0 ≤ i0 ∧ 0 ≤ j0 ∧ j0 < j ⊦ 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: zero-ge-add: ⊦ 0 ≤ j1 ∧ j1 < j → (0 ≤ i0 * j + j1 ↔ 0 ≤ i0);%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%abs-16: ⊦ 0 < j → abs(j) = j; from specification int-abs%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%lst-01: ⊦ ¬ i < j → (i ≠ j ↔ j < i); from specification int-basic1