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Proof step 2 from proof two-loup-01 (previous step)

m < n ⊦ ((m + n) / 2 - 1) + 1 = (m + n) / 2, m + 1 = n

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: le-div-to-mult: m0 ≠ 0 ⊦ n ≤ m / m0 ↔ n * m0 ≤ m;%n ≤ m ⊦ (m - n) + n0 = n1 ↔ m + n0 = n + n1; (s)% ⊦ ¬ m < n → ¬ m + m0 < n; (s)% ⊦ m < 1 ↔ m = 0; (s)% ⊦ 0 < m ↔ m ≠ 0; (s)%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic% ⊦ m < n → (m + n < 2 ↔ m = 0 ∧ n = 1); (ws)% ⊦ m ≤ n ↔ ¬ n < m; (ss)%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult% ⊦ m < n → n ≠ 0; (s)%succ-03: ⊦ m < n ∧ m + 1 ≠ n → m + 1 < n; (forward) from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic