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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
ints1 = [] ∨ (∀ ints0. ints0 = [] ∨ max(ints1 + ints0) = max(max(ints1), max(ints0))) ⊦ ints0 = [], max((i ' + ints1) + ints0) = max(max(i ' + ints1), max(ints0))

Rule argument

Additional information

Used simplifier rules: a: ⊦ (ints + ints0) + ints1 = ints + ints0 + ints1; from specification list[intlista]%e-01: ⊦ i ' + ints ≠ []; from specification list[intlista]%c: ⊦ i + ints = i ' + ints; from specification list[intlista]