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Proof step 14 from proof butmin-put-01 (previous step)

st[a, d0] ≠ ∅, a = st[a, d0].keys.min ⊦ a = st[a, d].min, a ∈ st, ¬ a ∈ st[a, (∅)[a]] -- st[a, (∅)[a]].min ∧ d = (∅)[a]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
a = (st.keys ++ a).min ⊦ a = st[a, d].min, a ∈ st, ¬ a ∈ st[a, (∅)[a]] -- st[a, (∅)[a]].min ∧ d = (∅)[a]

Rule argument

Additional information

Used simplifier rules: insert-03: ⊦ st[a, d].keys = st.keys ++ a;%neq: ⊦ st[a, d] ≠ ∅; from specification store