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Proof step 2 from proof ordered-app-max (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
a < x0.last, max(x0) < x0.last, ordered≤(x0) ⊦ x0 = [], a < max(x0)

Rule argument

Additional information

Used simplifier rules: nll: ⊦ ¬ b < a ∧ b < c → a < c; (forward) from specification oelem%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ ordered≤(x) ∧ x ≠ [] → (ordered≤(x + a ') ↔ ¬ a < x.last); (s)