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Proof step 23 from proof div-mult-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
0 < i0, i < 0, j < 0 ⊦ (i * i0) / (j * i0) = i / j, j = 0

Rule argument

Additional information

Used simplifier rules: forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs