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

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: s.min

Additional information

Used simplifier rules: min-less: ⊦ a ≠ s.min ∧ a ∈ s → s.min < a; Used terms: s.min