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Proof step 11 from proof square-ls-04 (previous step)

m < n, m * m + m0 < (m + 1) * (m + 1), m0 ≤ 2 * m, n * n ≤ m * m + m0 ⊦

User applied rule cut formula

And obtained the following new goals

Proof step 12:
m * m + m0 < (m + 1) * (m + 1), m < n, n * n < (m + 1) * (m + 1), n * n ≤ m * m + m0, m0 ≤ 2 * m ⊦

Rule argument

fma: n * n < (m + 1) * (m + 1)

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;%sqzero: ⊦ m * m = 0 ↔ m = 0;%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m ≤ n → (n < m ↔ false); (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat% ⊦ m < n → n ≠ 0; (s) Used formulas: n * n < (m + 1) * (m + 1)