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Proof step 3 from proof ls-03 (previous step)

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

User applied rule insert spec-lemma

And obtained the following new goals

Proof step 4:
⊦ i0 / i < i0 / j ∧ 0 ≤ i0 ∧ 0 < i ∧ 0 < j → j < i

Rule argument

spec: int-div name: lsf-01 seq: ⊦ i0 / i < i0 / j ∧ 0 ≤ i0 ∧ 0 < i ∧ 0 < j → j < i subst: [i0, i, j] → [i, 1, n→i(n)]

Additional information

Used simplifier rules: num-n2i-ls-02: 0 ≤ i ⊦ (* i) < n→i(n) ↔ i→n(i) < n;%nat2int-14: ⊦ n→i(n) < 1 ↔ n = 0;% ⊦ i / 1 = i; (s)% ⊦ 0 < n ↔ n ≠ 0; (s) Used terms: i, 1, n→i(n)