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Proof step 2 from proof square-ls-04 (previous step)

m0 ≤ 2 * m ⊦ m * m + m0 < n * n ↔ m < n

User applied rule case distinction

And obtained the following new goals

Proof step 3:
m * m + m0 < n * n, m0 ≤ 2 * m ⊦ m < n
Proof step 8:
m < n, m0 ≤ 2 * m ⊦ m * m + m0 < n * n

Rule argument

pos: right 1

Additional information