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

⊦ a ∈ st → (st \ (s ++ a)) -- a = st \ (s ++ a)

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: Diff-rec: ⊦ st \ (s ++ a) = (st \ s) -- a;%deldel: ⊦ st -- a -- a = st -- a; from specification store