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Proof step 12 from proof max-a (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
max(m, n) < m, n0 < m, n < m, n < n0, max(m, n) < n0, n + 1 < m, max(m, n) + 1 < m ⊦ m = n0

Rule argument

Additional information

Used simplifier rules: Max-left: ⊦ m < n → max(m, n) = n;% ⊦ m +1 = m + 1; (s)%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic%trans-01: ⊦ n < n0 ∧ n0 < n1 → n +1 < n1; (forward) from specification nat-basic% ⊦ m ≠ n → (m ≤ n ↔ m < n); (s)