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Proof step 3 from proof zero-05 (previous step)

i / j < i, i < 0 ⊦ j = 0

User applied rule insert lemma

And obtained the following new goals

Proof step 4:
j ≠ 0 ⊦ i < 0 ∧ j < 0 → 0 ≤ i / j
Proof step 5:
i < 0, i / j < i, i < 0 ∧ j < 0 → 0 ≤ i / j ⊦ j = 0

Rule argument

spec: <toplevel> name: zero-06 seq: j ≠ 0 ⊦ i < 0 ∧ j < 0 → 0 ≤ i / j subst: [j, i] → [j, i]

Additional information

Used terms: j, i