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Proof step 1 from proof Diff-mod (conclusion)

⊦ st[a, d] \ (s ++ a) = st \ (s ++ a)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ (st[a, d] \ s) -- a = (st \ s) -- a

Rule argument

Additional information

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