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Proof step 4 from proof square-ls-03 (previous step)

((n + 1) * (n + 2)) / 2 < ((m + 1) * (m + 2)) / 2 ↔ n < m, m ≤ n, m0 ≤ m ⊦ (m * (m + 1)) / 2 + m0 < ((n + 1) * (n + 2)) / 2

User applied rule case distinction

And obtained the following new goals

Proof step 5:
m0 ≤ m, m ≤ n ⊦ (m * (m + 1)) / 2 + m0 < ((n + 1) * (n + 2)) / 2, n < m, ((n + 1) * (n + 2)) / 2 < ((m + 1) * (m + 2)) / 2

Rule argument

pos: left 1

Additional information

Used simplifier rules: ⊦ m < n → (n ≤ m ↔ false); (s)