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Proof step 2 from proof zero-08 (previous step)

i < i / (n→i(n) + 1), i < 0, 0 < i / (n→i(n) + 1), i + 1 < i / (n→i(n) + 1) ⊦

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) + 1, i] flags:

Additional information

Used simplifier rules: nat2int: ⊦ 0 ≤ n→i(n);%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%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) + 1, i