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Proof step 2 from proof max-q (
previous step
)
a ∈ s, a < s.max, s.min < s.max, s.min ∈ s, s.max ∈ s, ∀ b. b ∈ s → ¬ a < b ⊦ s = ∅, # s = 0, 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 simplifier rules: max-in: ⊦ s ≠ ∅ → (¬ s.max ∈ s ↔ false); Used terms: s.max