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Proof step 6 from proof max-app (previous step)

ints1 + ints0 ≠ [], ints1 = [] ∨ (∀ ints0. ints0 = [] ∨ max(ints1 + ints0) = max(max(ints1), max(ints0))) ⊦ ints0 = [], max(i, max(ints1 + ints0)) = max(max(i ' + ints1), max(ints0))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
∀ ints0. ints0 = [] ∨ max(ints1 + ints0) = max(max(ints1), max(ints0)) ⊦ ints0 = [], max(i, max(ints1 + ints0)) = max(max(i ' + ints1), max(ints0)), ints1 + ints0 = [], ints1 = []

Rule argument

Additional information

Used simplifier rules: max-one: ⊦ max(i ') = i;%app: ⊦ ints + [] = ints; from specification list[intlista]%a: ⊦ (ints + ints0) + ints1 = ints + ints0 + ints1; from specification list[intlista]%Nil: ⊦ [] + ints = ints; from specification list[intlista]