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Proof step 2 from proof mult-ls-03 (previous step)

i0 < i / n→i(n) ⊦ n = 0, i0 * n→i(n) < i

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
i0 < i / n→i(n), i ≤ i0 * n→i(n) ⊦ n = 0

Rule argument

Additional information

Used simplifier rules: ⊦ i < j ↔ ¬ j ≤ i; (ss)