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Proof step 22 from proof less-mult-div-01 (previous step)

i1 < 0, 0 < i0 * j + j0, 0 < j, 0 < j1, i1 < j, j0 < j, 0 < (i1 * (i0 * j + j0)) / j, 0 ≤ (i1 * (i0 * j + j0)) / j, 0 ≤ i0 * j + j0, i0 * j + j0 ≤ (i1 * (i0 * j + j0)) / j, 0 ≤ i0, 0 ≤ j0 ⊦ j0 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
0 < (i1 * (i0 * j + j0)) / j, 0 < j, 1 < j, 0 < j0, i1 < 0, 0 < i0 * j + j0, 0 < j1, i1 < j, j0 < j, 0 ≤ (i1 * (i0 * j + j0)) / j, 0 ≤ i0 * j + j0, i0 * j + j0 ≤ (i1 * (i0 * j + j0)) / j, 0 ≤ i0 ⊦ j0 = 0

Rule argument

Additional information

Used simplifier rules: ⊦ i < 0 ∧ 0 ≤ j → i < j; (s)%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%forward-lsle: ⊦ 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% ⊦ i ≠ j → (i ≤ j ↔ i < j); (s)