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Proof step 1 from proof restrict-in (conclusion)

⊦ a ∈ st | s ↔ a ∈ st ∧ a ∈ s

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ a ∈ st \ (dom st \ s) ↔ a ∈ st ∧ a ∈ s

Rule argument

Additional information

Used simplifier rules: restrict: ⊦ st | s = st \ (dom st \ s);