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

s.min ∈ s, s.max ∈ s ⊦ (s ++ a).max = (a < s.max ⊃ s.max;a), s = ∅, # s = 0

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 4:
a < s.max, s.min ∈ s, s.max ∈ s ⊦ (s ++ a).max = s.max, s = ∅, # s = 0
Proof step 6:
¬ a < s.max, s.min ∈ s, s.max ∈ s ⊦ (s ++ a).max = a, s = ∅, # s = 0

Rule argument

pos: right 1

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