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Proof step 3 from proof zero-09 (previous step)

i < i / n→i(n), i < 0, 0 < i / n→i(n), i + 1 < i / n→i(n) ⊦ n = 0

User applied rule insert spec-lemma

And obtained the following new goals

Proof step 4:
⊦ 0 < j ∧ i < 0 → i / j ≤ 0

Rule argument

spec: int-div name: zero-01 seq: ⊦ 0 < j ∧ i < 0 → i / j ≤ 0 subst: [j, i] → [n→i(n), i]

Additional information

Used simplifier rules: num-n2i-ls-02: 0 ≤ i ⊦ (* i) < n→i(n) ↔ i→n(i) < n;% ⊦ 0 < n ↔ n ≠ 0; (s)% ⊦ i < j → (j ≤ i ↔ false); (as) Used terms: n→i(n), i