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

0 ≤ i, i / n→i(n) < 0 ⊦ n = 0

User applied rule insert spec-lemma

And obtained the following new goals

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

Rule argument

spec: int-div inst: name: zero-03 seq: ⊦ 0 < i ∧ 0 < j → 0 ≤ i / j subst: [i, j] → [i, n→i(n)] flags:

Additional information

Used simplifier rules: num-n2i-ls-02: 0 ≤ i ⊦ (* i) < n→i(n) ↔ i→n(i) < n;%nat2int-02: ⊦ n→i(n) = 0 ↔ n = 0;%zero: ⊦ 0 < m ↔ m ≠ 0; from specification nat-basic1%lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%anti: ⊦ i ≤ j → (j ≤ i ↔ i = j); from specification int-abs%zero: j ≠ 0 ⊦ 0 / j = 0; from specification int-div Used terms: i, n→i(n)