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Proof step 7 from proof butmin-put-01 (previous step)

¬ a0 ≠ st[a, d].min, a = st[a, d0].min ⊦ a ∈ st, (a0 ∈ st[a, d] -- st[a, d].min ↔ a0 ∈ st) ∧ (st[a, d] -- st[a, d].min)[a0] = st[a0]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
a = st[a, d0].min ⊦ a ∈ st, (st[a, d].min ∈ st[a, d] -- st[a, d].min ↔ st[a, d].min ∈ st) ∧ (∅)[st[a, d].min] = st[st[a, d].min]

Rule argument

Additional information

Used simplifier rules: lookupisempty: ⊦ (st -- a)[a] = (∅)[a]; from specification store