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Proof step 9 from proof square-ls-04 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
m < n, n * n ≤ m * m + m0, m0 ≤ 2 * m ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m < n ↔ ¬ n ≤ m; (ss)