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Proof step 10 from proof square-ls (previous step)

3 * m0 + m0 * m0 < 3 * m + m * m, m ≤ m0, m * m + m ≤ m0 * m0 + m0 ⊦ 3 * m + m * m = 0

User applied rule cut formula

And obtained the following new goals

Proof step 11:
3 * m + m * m ≤ 3 * m0 + m * m, 3 * m0 + m0 * m0 < 3 * m + m * m, m ≤ m0, m * m + m ≤ m0 * m0 + m0 ⊦ 3 * m + m * m = 0

Rule argument

fma: 3 * m + m * m ≤ 3 * m0 + m * m

Additional information

Used simplifier rules: le-num: 0 ≠ m ⊦ (* m) * n ≤ m * n0 ↔ n ≤ n0;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%lep-02: ⊦ m + n ≤ m + n0 ↔ n ≤ n0; from specification nat%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%le-ls-simp: ⊦ m ≤ n → (n < m ↔ false); from specification nat%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used formulas: 3 * m + m * m ≤ 3 * m0 + m * m