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Proof step 1 from proof min-in (conclusion)

⊦ st.min = a ∧ st ≠ ∅ → a ∈ st

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: min-def: ⊦ st ≠ ∅ → st.min = st.keys.min;%mk-in: ⊦ st ≠ ∅ → (¬ st.keys.min ∈ st ↔ false);