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Proof step 1 from proof twoeqf-02 (conclusion)

⊦ m + m0 = 2 * n + 1 ∧ n < m → m0 ≤ n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat