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

⊦ max(ins≤(a, x)) = a, a < max(x)

Heuristic structural induction applied rule structural induction

And obtained the following new goals

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

Rule argument

term: x

Additional information

Used simplifier rules: max-base: ⊦ max(a ') = a;%list freely generated by [], +;;% ⊦ ins≤(a, []) = a '; (s)