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Proof step 6 from proof div-eq-zero (previous step)

i < j, 0 < i, 0 < j ⊦ i / j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
i < j, 0 < i, 1 < j, 0 < j ⊦ i / j = 0

Rule argument

Additional information

Used simplifier rules: forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1