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Proof step 5 from proof two-loup-11 (previous step)

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

User applied rule cut formula

And closed the goal!

Rule argument

fma: n0 = 0

Additional information

Used simplifier rules: twols-02: ⊦ m ≠ 0 → (m < 2 ↔ m = 1); from specification nat-basic1%twoeqf-02: ⊦ m + m0 = 2 * n + 1 ∧ n < m → m0 ≤ n; (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-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%anti: ⊦ m ≤ n → (n ≤ m ↔ m = n); from specification nat%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic1%numeral-19: m2 ≠ 0 ⊦ (* m2) * m + n = m + n0 ↔ (m2 - 1) * m + n = n0; from specification nat-mult%one: ⊦ m0 * 1 = m0; from specification nat-mult%add-base: ⊦ n + 0 = n; from specification nat-basic2%twoeqf: ⊦ m + m0 = 2 * n ∧ n < m → m0 < n; (forward) from specification nat-mult%le-ls-simp: ⊦ m ≤ n → (n < m ↔ false); from specification nat Used formulas: n0 = 0