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

j < 0, i < i0, i < 0, 0 < i0, i + 1 < i0 ⊦ (i * i0) / (j * i0) = i / j, j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
j < i0, j < 0, i + 1 < i0, 0 < i0, i < 0, i < i0, j + 1 < i0 ⊦ j = 0, (i * i0) / (j * i0) = i / j

Rule argument

Additional information

Used simplifier rules: succ-01: ⊦ i +1 = i + 1; from specification int-basic2%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1