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Proof step 5 from proof le-03 (previous step)

⊦ m +1 * n ≤ m +1 * n0 ↔ n ≤ n0

User applied rule apply lemma

And obtained the following new goals

Proof step 6:
⊦ m0 ≠ 0 → (m0 * m ≤ m0 * n ↔ m ≤ n)
Proof step 7:
m +1 ≠ 0 → (m +1 * n ≤ m +1 * n0 ↔ n ≤ n0) ⊦ m +1 * n ≤ m +1 * n0 ↔ n ≤ n0

Rule argument

spec: <toplevel> name: le seq: ⊦ m0 ≠ 0 → (m0 * m ≤ m0 * n ↔ m ≤ n) subst: [m0, m, n] → [m +1, n, n0]

Additional information

Used terms: m +1, n, n0