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Proof step 2 from proof dom-add-in (previous step)

a ∈ st ⊦ dom st ++ a = dom st

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: Dom: ⊦ a ∈ dom st ↔ a ∈ st;% ⊦ s ++ a = s ↔ a ∈ s; (s)