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Proof step 2 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 3:
⊦ 0 < j ∧ i < 0 → i / j ≤ 0

Rule argument

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

Additional information

Used simplifier rules: num-n2i-ls-02: 0 ≤ i ⊦ (* i) < n→i(n) ↔ i→n(i) < n;%zero: ⊦ 0 < m ↔ m ≠ 0; from specification nat-basic1%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs Used terms: n→i(n), i