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Proof step 1 from proof ordered-app-max (conclusion)

⊦ x0 ≠ [] ∧ ¬ a < max(x0) ∧ ordered≤(x0) → (ordered≤(x0 + a ') ↔ true)

System applied rule normalize

And obtained the following new goals

Proof step 2:
ordered≤(x0) ⊦ ordered≤(x0 + a ') ↔ true, x0 = [], a < max(x0)

Rule argument

Additional information