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

0 < i ⊦ i * j ≤ 0 ↔ j ≤ 0

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 * j ≤ i * 0 ↔ j ≤ 0) ⊦ i * j ≤ 0 ↔ j ≤ 0

Rule argument

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

Additional information

Used terms: i, j, 0