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Proof step 7 from proof square-ls (previous step)

m0 * m0 + m0 < m * m + m ↔ m0 + 1 < n, n = m + 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 8:
m0 * m0 + m0 < m * m + m ↔ m0 < m ⊦ 3 * m0 + m0 * m0 < 3 * m + m * m ↔ m0 < m

Rule argument

Additional information

Used simplifier rules: dist: ⊦ m0 * (m + n) = m0 * m + m0 * n;%one: ⊦ m0 * 1 = m0;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%two-02: ⊦ m + m = 2 * m;%succ: ⊦ (* m0) * m + m = (m0 + 1) * m;% ⊦ m + 1 ≠ 0; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n < m + n0 ↔ n < n0; (s)