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Proof step 2 from proof twoeqf-02 (previous step)

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

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;%nat freely generated by 0, +1;;%succ: ⊦ m +1 = m + 1; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%le-ls-simp: ⊦ m ≤ n → (n < m ↔ false); from specification nat