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Proof step 1 from proof two-loup-11 (conclusion)

⊦ m < n → ((m + n) / 2 < m + 1 ↔ m + 1 = n)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: two-mdiv2-03: ⊦ (2 * m + 1) / 2 = m;%succ-03: ⊦ m < n ∧ m + 1 ≠ n → m + 1 < n; (forward) from specification nat-basic2%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%ref-1: ⊦ ¬ n < n; from specification nat-basic1%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%two-02: ⊦ m + m = 2 * m; from specification nat-mult%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2