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Proof step 2 from proof twoeqf-02 (previous step)

m + m0 = 2 * n + 1, n < m ⊦ m0 ≤ n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
m + m0 = 2 * n + 1, n < m0, n < m ⊦

Rule argument

Additional information

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