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Proof step 5 from proof union-left-empty (previous step)

⊦ (a ∈ st ⊃ st[a];(∅)[a]) = st[a]

Heuristic if-then-else split applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: ⊦ ¬ a ∈ st → st[a] = (∅)[a]; (s)