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Proof step 1 from proof diff-add-notin (conclusion)

⊦ ¬ a ∈ st → st \ (s ++ a) = st \ s

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ (st \ s) -- a = st \ s, a ∈ st

Rule argument

Additional information

Used simplifier rules: Diff-rec: ⊦ st \ (s ++ a) = (st \ s) -- a;