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Proof step 19 from proof div-mult-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
0 < i0 ⊦ (i * i0) / (j * i0) = i / j, j = 0

Rule argument

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs% ⊦ i < j → ¬ j < i; (s)% ⊦ ¬ i < j → (i = j ↔ ¬ j < i); (ss)