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Proof step 1 from proof empty-02 (conclusion)

⊦ a ∈ st → (¬ st.keys ≠ ∅ ↔ false)

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: Min-min: ⊦ a ∈ st → ¬ a < st.min;%inf: ⊦ a ∈ st → a ∈ st.keys;%notempty: ⊦ a ∈ st.keys → st.keys ≠ ∅;