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Proof step 16 from proof max-a (previous step)

max(n, n0) ≤ m ⊦ m = max(max(m, n), n0), n < n0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
n ≤ m, n0 ≤ n, n0 ≤ m ⊦ m = max(max(m, n), n0)

Rule argument

Additional information

Used simplifier rules: Max-right: ⊦ ¬ m < n → max(m, n) = m;%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat% ⊦ m < n ↔ ¬ n ≤ m; (ss)