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Proof step 4 from proof exdivmodpos (previous step)

i0 * j ≤ i ∧ i < (i0 + 1) * j ∧ 0 ≤ i0 ⊦ 0 ≤ i ∧ 0 < j → (∃ i0, j0. i = i0 * j + j0 ∧ 0 ≤ i0 ∧ 0 ≤ j0 ∧ j0 < j)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: zero-gr: ⊦ 0 < i → (0 < i * j ↔ 0 < j);%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%ls-pos: ⊦ 0 < i → (i * j < i * j0 ↔ j < j0);%ler: ⊦ i ≤ i; from specification int-abs%succ-02: ⊦ i < j + 1 ↔ i ≤ j; from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%mone-06: ⊦ ~1 < i ↔ 0 ≤ i; from specification int-abs%numls-02: ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; from specification int-abs%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs