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Proof step 1 from proof two-loup-05 (conclusion)

⊦ m < n → ¬ (m + n) / 2 + 1 < m

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2%trans-1: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic1%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1