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Proof step 1 from proof square-less (conclusion)

⊦ m * m < n * n ↔ m < n

User applied rule apply lemma

And obtained the following new goals

Proof step 2:
⊦ n < n0 ∧ n0 < n1 → n < n1
Proof step 3:
m * m < m * n ∧ m * n < n * n → m * m < n * n ⊦ m * m < n * n ↔ m < n

Rule argument

spec: nat-basic name: trans seq: ⊦ n < n0 ∧ n0 < n1 → n < n1 subst: [n0, n, n1] → [m * n, m * m, n * n]

Additional information

Used terms: m * n, m * m, n * n