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Proof step 1 from proof max-ifins (conclusion)

⊦ (s ++ a).max = (s = ∅ ⊃ a;(a < s.max ⊃ s.max;a))

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 2:
s ≠ ∅ ⊦ (s ++ a).max = (a < s.max ⊃ s.max;a)

Rule argument

pos: right 1

Additional information

Used simplifier rules: max-one: ⊦ (∅ ++ a).max = a;