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Proof step 3 from proof twoeqf-02 (previous step)

m + m0 = 2 * n + 1, n < m0, n < m ⊦

Heuristic structural induction applied rule structural induction

And closed the goal!

Rule argument

term: m0

Additional information

Used simplifier rules: twoeqf: ⊦ m + m0 = 2 * n ∧ n < m → m0 < 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;%nat freely generated by 0, +1;;% ⊦ m +1 = m + 1; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n = m + n0 ↔ n = n0; (s)%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%trans-01: ⊦ n < n0 ∧ n0 < n1 → n +1 < n1; (forward) from specification nat-basic%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic% ⊦ m < n → n ≠ 0; (s)% ⊦ ¬ n < n; (s)% ⊦ 0 + m = m; (s)% ⊦ ¬ n < 0; (s)