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Proof step 2 from proof zero-12 (previous step)

i / n→i(n) < 0, 0 ≤ i ⊦ n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
i / n→i(n) < 0, i / n→i(n) < i, 0 ≤ i ⊦ n = 0

Rule argument

Additional information

Used simplifier rules: ⊦ ~1 < i ↔ 0 ≤ i; (s)%numlef-01: ⊦ (* i) ≤ j → i - 1 < j; (forward) from specification int-abs%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs