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Proof step 3 from proof max-ins (previous step)

⊦ max(ins≤(a, x)) = max(x), x = [], max(x) < a

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 4:
∀ a. max(ins≤(a, x0)) = max(x0) ∨ x0 = [] ∨ max(x0) < a ⊦ a0 + x0 = [], max(ins≤(a, a0 + x0)) = max(a0 + x0), max(a0 + x0) < a

Rule argument

term: x

Additional information

Used simplifier rules: list freely generated by [], +;;