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Proof step 3 from proof ordered-app-max (previous step)

a < x0.last, max(x0) < x0.last, ordered≤(x0) ⊦ x0 = [], a < max(x0)

User applied rule structural induction

And obtained the following new goals

Proof step 4:
max(a0 + x) < (a0 + x).last, a < (a0 + x).last, ordered≤(a0 + x), ∀ a. a < x.last ∧ max(x) < x.last ∧ ordered≤(x) → x = [] ∨ a < max(x) ⊦ a0 + x = [], a < max(a0 + x)

Rule argument

indvar: x0

Additional information

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