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Proof step 2 from proof square-ls (previous step)

∀ n. m0 * (m0 + 1) < n * (n + 1) ↔ m0 < n ⊦ m0 +1 * (m0 +1 + 1) < n * (n + 1) ↔ m0 +1 < n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ n0. m0 * m0 + m0 < n0 * n0 + n0 ↔ m0 < n0 ⊦ 3 * m0 + m0 * m0 + 2 < n * n + n ↔ m0 + 1 < n

Rule argument

Additional information

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