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Proof step 5 from proof square-ls (previous step)

m0 * m0 + m0 < n -1 * n -1 + n -1 ↔ m0 < n -1 ⊦ 3 * m0 + m0 * m0 + 2 < n * n + n ↔ m0 + 1 < n, n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
m0 * m0 + m0 < (n - 1) * (n - 1) + (n - 1) ↔ m0 + 1 < n ⊦ 3 * m0 + m0 * m0 + 2 < n * n + n ↔ m0 + 1 < n, n = 0

Rule argument

Additional information

Used simplifier rules: n0 ≤ n ⊦ m < n - n0 ↔ m + n0 < n; (s)% ⊦ m < 1 ↔ m = 0; (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ n -1 = n - 1; (s)