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Proof step 3 from proof butmin-put (previous step)

⊦ st[st.min, d] -- st[st.min, d].min = st, st.min ∈ st

Heuristic weak cut applied rule cut formula

And closed the goal!

Rule argument

fma: st = ∅

Additional information

Used simplifier rules: min: ⊦ (∅)[a, d].min = a;%min-def: ⊦ st ≠ ∅ → st.min = st.keys.min;%mk-in: ⊦ st ≠ ∅ → (st.keys.min ∈ st ↔ true);% ⊦ st[a, d] -- a = st -- a; (s)% ⊦ ∅ -- a = ∅; (s) Used formulas: st = ∅