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Proof step 8 from proof butmin-put-01 (previous step)

a = st[a, d0].min, a0 = st[a, d].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 9:
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: ⊦ (st -- a)[a] = (∅)[a]; (s)