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Proof step 2 from proof two-loup-11 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
(m + n) / 2 < n, m + 1 < n, m < n, (m + n) / 2 ≤ m ⊦ m + 1 = n

Rule argument

Additional information

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