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Proof step 1 from proof butmin-in (conclusion)

⊦ a ∈ st.butmin ↔ st ≠ ∅ ∧ a ∈ st ∧ a ≠ st.min

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

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