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Proof step 2 from proof sqr-09 (previous step)

2 < m, n * n ≤ m + 1, m ≤ n ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
2 < m, 2 < n, m ≤ n, n * n ≤ m + 1 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic