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Proof step 9 from proof butmin-put-01 (previous step)

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]

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 2

Additional information

Used simplifier rules: min-put-01: ⊦ st[a, d].min = a → st[a, d0].min = a;%min-put-01: ⊦ st[a, d].min = a → st[a, d0].min = a;% ⊦ ¬ a ∈ st → st[a] = (∅)[a]; (s)% ⊦ st[a, d] -- a = st -- a; (s)% ⊦ ¬ a ∈ st → st -- a = st; (s)