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Proof step 5 from proof two-loup-07 (previous step)

m + n = m0 * 2 + n0, n0 < 2, n + 1 < m0, m < n ⊦ m0 = 0, n = 0

User applied rule cut formula

And obtained the following new goals

Proof step 6:
m + n = m0 * 2 + n0, n0 < 2, n + 1 < m0, m < n ⊦ m + n < 2 * m0, m0 = 0, n = 0

Rule argument

fma: m + n < 2 * m0

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic1%zero-06: ⊦ m ≠ 0 → (m * n = 0 ↔ n = 0); from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%modls: ⊦ n0 < m0 → (m0 * m + n0 < m0 * n ↔ m < n); from specification nat-mult%ref-1: ⊦ ¬ n < n; from specification nat-basic1%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used formulas: m + n < 2 * m0