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Proof step 2 from proof bmin-sub-04 (previous step)

s ⊆ s0 ⊦ s ⊆ s0.butmin ↔ ¬ s0.min ∈ s, s0 = ∅

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: Butmin: ⊦ s ≠ ∅ → s.butmin = s -- s.min;% ⊦ s ⊆ s0 -- a ↔ s ⊆ s0 ∧ ¬ a ∈ s; (ws)% ⊦ s ⊆ s0 ∧ ¬ a ∈ s → s ⊆ s0 -- a; (s)