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Proof step 2 from proof max-q (previous step)

a < s.max, s.min < s.max, a ∈ s, s.min ∈ s, s.max ∈ s, ∀ b. b ∈ s → ¬ a < b ⊦ # s = 0, s = ∅, s.max < s.min, a < s.min

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 6 terms: s.max

Additional information

Used terms: s.max