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Proof step 2 from proof ls-rm (previous step)

b0 ∈ s, s.min ∈ s, s.max ∈ s, ∀ b0. b0 ∈ s → b0 < b ⊦ s = ∅, b0 = a, # s = 0, b0 < s.min, s.max < b0, b0 < b, s.max < b

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: s.max

Additional information

Used simplifier rules: max-in: ⊦ s ≠ ∅ → (¬ s.max ∈ s ↔ false); Used terms: s.max