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Proof step 1 from proof two-loup-08 (conclusion)

⊦ m < n ∧ n < n0 → (m + n) / 2 < n0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
m < (m + n) / 2, m + 1 < n0, n < (m + n) / 2, m < n, n < n0, n0 ≤ (m + n) / 2, m < n0, m + 1 < (m + n) / 2 ⊦ m + n = 0, m = 0

Rule argument

Additional information

Used simplifier rules: two-zero-01-01: ⊦ m / 2 = 0 ↔ ¬ ¬(m = 0 ∨ m = 1);%two-ls-02: ⊦ ¬ m < m / 2;%z: ⊦ 0 < n ↔ n ≠ 0; from specification nat-basic1%z: ⊦ 0 + m = m; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%one: ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2%f: ⊦ m < n → n ≠ 0; from specification nat-basic1%trans-1: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic1%trans: ⊦ n < n0 ∧ n0 < n1 → n +1 < n1; (forward) from specification nat-basic1%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat