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Proof step 2 from proof butmin-id (previous step)

⊦ st.butmin[st.min, st[st.min]] = st, st = ∅

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-in: ⊦ st ≠ ∅ → st.min ∈ st;% ⊦ a ∈ st → (st -- a)[a, st[a]] = st; (s)