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Proof step 9 from proof empty-01 (previous step)

st.min ∈ st, ∀ a. a ∈ st.keys ↔ a ∈ ∅ ⊦ st = ∅, ¬ a ∈ st ∧ st[a] = (∅)[a]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
st.min ∈ st, ∀ a0. ¬ a0 ∈ st ⊦ st = ∅, ¬ a ∈ st ∧ st[a] = (∅)[a]

Rule argument

Additional information

Used simplifier rules: Keys: ⊦ a ∈ st.keys ↔ a ∈ st;% ⊦ ¬ a ∈ ∅; (s)