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Proof step 3 from proof At-in-out (previous step)

b ∈ st ⊦ b = a, st[a, d][b] = st[b], a ∈ st

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
b ∈ st ⊦ st[a, d][b] = st[b], a ∈ st

Rule argument

Additional information