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Proof step 3 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 4:
⊦ 0 < j ∧ i < 0 → i / j ≤ 0

Rule argument

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

Additional information

Used simplifier rules: nat2int: ⊦ 0 ≤ n→i(n);% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; (s)% ⊦ ~1 < i ↔ 0 ≤ i; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ i < j → (j ≤ i ↔ false); (as) Used terms: n→i(n) + 1, i