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Proof step 2 from proof zero-10 (previous step)

i / (n→i(n) + 1) < 0, 0 ≤ i ⊦

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) + 1] flags:

Additional information

Used simplifier rules: nat2int: ⊦ 0 ≤ n→i(n);%nat2int-07: ⊦ 0 ≠ n→i(m) + 1;%numls-02: ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; from specification int-abs%mone-06: ⊦ ~1 < i ↔ 0 ≤ i; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%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) + 1