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Proof step 3 from proof ls-same-01 (previous step)

0 < i ⊦ i < i * j ↔ 1 < j

User applied rule apply lemma

And obtained the following new goals

Proof step 4:
⊦ 0 < i → (i * j < i * j0 ↔ j < j0)
Proof step 5:
0 < i, 0 < i → (i * 1 < i * j ↔ 1 < j) ⊦ i < i * j ↔ 1 < j

Rule argument

spec: <toplevel> name: ls-pos seq: ⊦ 0 < i → (i * j < i * j0 ↔ j < j0) subst: [i, j, j0] → [i, 1, j]

Additional information

Used terms: i, 1, j