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Proof step 13 from proof max-app (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 14:
⊦ ints ≠ [] → max(i ' + ints) = max(i, max(ints))

Rule argument

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

Additional information

Used simplifier rules: max-a: ⊦ max(i, max(j, j0)) = max(max(i, j), j0); from specification int-min%max-c: ⊦ max(i, j) = max(j, i); from specification int-min Used terms: ints1, i