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Proof step 6 from proof two-loup-08 (previous step)

m + n = m0 * 2 + n1, n < m0, m < n0, m < n, n < n0, m + 1 < n0, n1 < 2, m < m0, m + 1 < m0, n0 ≤ m0 ⊦ m + n = 0, m = 0

User applied rule cut formula

And closed the goal!

Rule argument

fma: n1 = 0

Additional information

Used simplifier rules: ⊦ m ≠ 0 → (m < 2 ↔ m = 1); (s)%twoeqf-03: ⊦ m + m0 = 2 * n + 1 ∧ m < n → n + 1 < m0; (forward) 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%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m < n → n ≠ 0; (s)% ⊦ m + m0 < n → ¬ n + n0 < m; (s)% ⊦ 0 + m = m; (s)%twoeqf-01: ⊦ m + m0 = 2 * n ∧ m < n → n < m0; (forward) from specification nat-mult% ⊦ 2 * m = 0 ↔ m = 0; (s)% ⊦ m < n → ¬ n < m; (s) Used formulas: n1 = 0