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Proof step 4 from proof fle (previous step)

¬ m < n0 ∧ ¬ n < m → ¬ n < n0, m < n, n ≤ n0, n0 ≤ m ⊦ n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
m < n, n ≤ n0, n0 ≤ m, ¬ m < n0 → ¬ n < n0 ⊦ n = 0

Rule argument

Additional information

Used simplifier rules: ls-02: ⊦ m < n → ¬ n < m; from specification nat-basic1