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Proof step 2 from proof bmin-nin (previous step)

a ∈ s.butmin ⊦ s = ∅, a ∈ s

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: Butmin: ⊦ s ≠ ∅ → s.butmin = s -- s.min;% ⊦ a ∈ s -- b ↔ a ≠ b ∧ a ∈ s; (ws)