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Proof step 3 from proof auxexdivmod-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: zero-left: ⊦ 0 * i = 0;%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%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%lels: ⊦ ¬ i < j ↔ j ≤ i; 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%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs