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

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

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

Rule argument

pos: right 2

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