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Proof step 4 from proof max-app (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ ints ≠ [] → max(i ' + ints) = max(i, max(ints))
Proof step 6:
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))

Rule argument

spec: inst: name: max-rec seq: ⊦ ints ≠ [] → max(i ' + ints) = max(i, max(ints)) subst: [ints, i] → [ints1 + ints0, i] flags: rewrite, current

Additional information

Used simplifier rules: app-02-ant: ⊦ ¬ ints + ints0 ≠ [] ↔ ints = [] ∧ ints0 = []; from specification list[intlista]%a: ⊦ (ints + ints0) + ints1 = ints + ints0 + ints1; from specification list[intlista] Used terms: ints1 + ints0, i