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Proof step 2 from proof zero-09 (previous step)

i < 0 ⊦ n = 0, i / n→i(n) ≤ 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
i < i / n→i(n), i < 0, 0 < i / n→i(n), i + 1 < i / n→i(n) ⊦ n = 0

Rule argument

Additional information

Used simplifier rules: ⊦ i +1 = i + 1; (s)% ⊦ 1 ≤ i ↔ 0 < i; (s)%trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%numlsf-02: ⊦ (* i) < j → i + 1 ≤ j; (forward) from specification int-abs% ⊦ i ≤ j ↔ ¬ j < i; (ss)