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Proof step 1 from proof two-loup-02 (conclusion)

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

Rule argument

Additional information

Used simplifier rules: 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%succ-03: ⊦ m < n ∧ m + 1 ≠ n → m + 1 < n; (forward) from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1