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Proof step 4 from proof square-two (previous step)

n * n + n = m * 2, m ≤ n ⊦ n = m ∧ m < 2

User applied rule cut formula

And obtained the following new goals

Proof step 5:
n * n + n = m * 2, m ≤ n, m * 2 ≤ n * 2 ⊦ n = m ∧ m < 2

Rule argument

fma: m * 2 ≤ n * 2

Additional information

Used simplifier rules: ls: ⊦ m0 * m < m0 * n ↔ m0 ≠ 0 ∧ m < n;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%twoeqf-01: ⊦ m + m0 = 2 * n ∧ m < n → n < m0;% ⊦ m ≤ n ↔ ¬ n < m; (ss)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m < n → n ≠ 0; (s)% ⊦ ¬ n < n; (s) Used formulas: m * 2 ≤ n * 2