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Proof step 1 from proof twoeqf-03 (conclusion)

⊦ m + m0 = 2 * n + 1 ∧ m < n → n + 1 < m0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
m + m0 = 2 * n + 1, m < n, m0 ≤ n + 1 ⊦ n = 0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat