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Proof step 2 from proof two-loup-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: ⊦ m +1 = m + 1; (s)% ⊦ 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%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic%trans-01: ⊦ n < n0 ∧ n0 < n1 → n +1 < n1; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic