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Proof step 6 from proof max-ifins (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
s.min ∈ s, s.max ∈ s ⊦ # s = 0, (s ++ a).max = a, s = ∅, a < s.max, s.max < s.min, a < s.min

Rule argument

Additional information

Used simplifier rules: nnn: ⊦ ¬ c < b ∧ ¬ a < c → ¬ a < b; (forward) from specification oelem