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Proof step 3 from proof square-interval (previous step)

n * (n + 1) + 2 * n0 < (m + 1) * (m + 2), n * (n + 1) + 2 * n0 < (n + 1) * (n + 2), m * (m + 1) ≤ n * (n + 1) + 2 * n0, n0 ≤ n ⊦ m = n

User applied rule cut formula

And obtained the following new goals

Rule argument

fma: m * (m + 1) < (n + 1) * ((n + 1) + 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;%succ: ⊦ (* m0) * m + m = (m0 + 1) * m;%numeral-24: m1 ≠ 0 ⊦ m + n < (* m1) * m + n0 ↔ n < (m1 - 1) * m + n0;%coef-21: ⊦ m * n < m1 * n + n ↔ m < m1 + 1 ∧ n ≠ 0;%numeral-20: m2 ≠ 0 ⊦ (* m2) * m + n ≤ m + n0 ↔ (m2 - 1) * m + n ≤ n0;%coef-18: ⊦ m * n + n ≤ m1 * n ↔ m + 1 ≤ m1 ∨ n = 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; (ss)%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%n < n0 ⊦ (* n) ≠ m0 + (* n0); (s)% ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)% ⊦ m + 1 ≤ n ↔ m < n; (s)%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat% ⊦ m < n → n ≠ 0; (s)% ⊦ ¬ n < n; (s) Used formulas: m * (m + 1) < (n + 1) * ((n + 1) + 1)