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Proof step 8 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 9:
m < n, m0 ≤ 2 * m, n * n ≤ m * m + m0 ⊦ n = 0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat