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

i ≤ 0 ⊦ i * j < 0 ↔ 0 < j ∧ i ≠ 0

User applied rule apply lemma

And obtained the following new goals

Proof step 4:
⊦ i < 0 → (i * j < i * j0 ↔ j0 < j)
Proof step 5:
i ≤ 0, i < 0 → (i * j < i * 0 ↔ 0 < j) ⊦ i * j < 0 ↔ 0 < j ∧ i ≠ 0

Rule argument

spec: <toplevel> name: ls-neg seq: ⊦ i < 0 → (i * j < i * j0 ↔ j0 < j) subst: [i, j, j0] → [i, j, 0]

Additional information

Used terms: i, j, 0