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Proof step 14 from proof square-ls (previous step)

m0 < m, m0 * m0 + m0 < m * m + m ↔ m0 < m ⊦ 3 * m0 + m0 * m0 < 3 * m + m * m

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
m0 < m, m0 * m0 + m0 < m * m + m, 3 * m + m * m ≤ 3 * m0 + m0 * m0 ⊦ m = 0, m * m + m = 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