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Proof step 1 from proof two-loup-04 (conclusion)

⊦ m < n ∧ m + 1 ≠ n → ¬ n < (m + n) / 2 - 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
2 + m < (m + n) / 2, m + 1 < n, m + 1 < (m + n) / 2, m < n, n + 1 < (m + n) / 2 ⊦ n = 0, (m + n) / 2 - 1 = 0, m + 1 = n, (m + n) / 2 = 0

Rule argument

Additional information

Used simplifier rules: le-div-to-mult: m0 ≠ 0 ⊦ n ≤ m / m0 ↔ n * m0 ≤ m;%zero-01: ⊦ m < n → m / n = 0;%ls: n0 ≤ n ⊦ m < n - n0 ↔ m + n0 < n; from specification nat%nls-fst: ⊦ ¬ m < n → ¬ m + m0 < n; from specification nat-basic2%one-01: ⊦ m < 1 ↔ m = 0; from specification nat-basic1%zero: ⊦ 0 < m ↔ m ≠ 0; from specification nat-basic1%one: ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); from specification nat-basic1%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic1%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic1%twols: ⊦ m < n → (m + n < 2 ↔ m = 0 ∧ n = 1); from specification nat-basic2%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%succ: ⊦ m +1 = m + 1; from specification nat-basic2%trans-1: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic1%succ-03: ⊦ m < n ∧ m + 1 ≠ n → m + 1 < n; (forward) from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1