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Proof step 3 from proof mod-zero-02 (previous step)

i % n→i(n) < 0, i % n→i(n) < i, 0 ≤ i ⊦ n = 0

User applied rule insert spec-lemma

And obtained the following new goals

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

Rule argument

spec: int-div name: mod-zero-01 seq: ⊦ 0 < j ∧ 0 ≤ i → ¬ 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) Used terms: n→i(n), i