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Proof step 5 from proof min-def (previous step)

st.min ∈ st, st.keys.min ∈ st → ¬ st.keys.min < st.min ⊦ st.min = st.keys.min, st = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
st.min < st.keys.min, st.min ∈ st ⊦ st.min = st.keys.min, st = ∅

Rule argument

Additional information

Used simplifier rules: mk-in: ⊦ st ≠ ∅ → (st.keys.min ∈ st ↔ true);% ⊦ a ≠ b → (a < b ↔ ¬ b < a); (ss)