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Proof step 10 from proof two-loup-04 (previous step)

m + n = (2 * m1 + 2) + n0, m < m1, n0 < 2, m + 1 < n, m < n, n < m1, 2 + m < m1, m + 1 < m1 ⊦ m + 1 = n

User applied rule cut formula

And obtained the following new goals

Proof step 11:
m + n = (2 * m1 + 2) + n0, m + 1 < m1, 2 + m < m1, n < m1, m < n, m + 1 < n, n0 < 2, m < m1 ⊦ m + 1 = n, m + n < 2 * m1

Rule argument

fma: m + n < 2 * m1

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic% ⊦ 2 * m = 0 ↔ m = 0; (s)% ⊦ m < n → n ≠ 0; (s)%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult% ⊦ ¬ m + n < m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used formulas: m + n < 2 * m1