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

0 < i ⊦ 0 ≤ i * j ↔ 0 ≤ j

User applied rule apply lemma

And obtained the following new goals

Proof step 4:
⊦ 0 < i → (i * j ≤ i * i0 ↔ j ≤ i0)
Proof step 5:
0 < i, 0 < i → (i * 0 ≤ i * j ↔ 0 ≤ j) ⊦ 0 ≤ i * j ↔ 0 ≤ j

Rule argument

spec: <toplevel> name: le-pos seq: ⊦ 0 < i → (i * j ≤ i * i0 ↔ j ≤ i0) subst: [i, j, i0] → [i, 0, j]

Additional information

Used terms: i, 0, j