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

ordered≤(x) ⊦ max(x) = x.last, x = []

User applied rule structural induction

And obtained the following new goals

Proof step 4:
ordered≤(a + x0), ordered≤(x0) → max(x0) = x0.last ∨ x0 = [] ⊦ a + x0 = [], max(a + x0) = (a + x0).last

Rule argument

indvar: x

Additional information

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