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Proof step 1 from proof square-ls-03 (conclusion)

⊦ m0 ≤ m ∧ m ≤ n → (m * (m + 1)) / 2 + m0 < ((n + 1) * (n + 2)) / 2

User applied rule insert lemma

And obtained the following new goals

Proof step 2:
⊦ ((m + 1) * (m + 2)) / 2 < ((n + 1) * (n + 2)) / 2 ↔ m < n
Proof step 3:
((n + 1) * (n + 2)) / 2 < ((m + 1) * (m + 2)) / 2 ↔ n < m ⊦ m0 ≤ m ∧ m ≤ n → (m * (m + 1)) / 2 + m0 < ((n + 1) * (n + 2)) / 2

Rule argument

spec: nat-div inst: name: square-ls seq: ⊦ ((m + 1) * (m + 2)) / 2 < ((n + 1) * (n + 2)) / 2 ↔ m < n subst: [m, n] → [n, m] flags: current, []

Additional information

Used terms: n, m