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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;%Min-min: ⊦ a ∈ st → ¬ a < st.min;%inf: ⊦ a ∈ st → a ∈ st.keys;%Keys: ⊦ a ∈ st.keys ↔ a ∈ st;%In-empty: ⊦ ¬ a ∈ ∅; from specification store%sls: ⊦ a ≠ b → (¬ a < b ↔ b < a); from specification oelem%Del-in: ⊦ a ∈ st -- b ↔ a ≠ b ∧ a ∈ st; from specification store