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Proof step 6 from proof Sep-eq (
previous step
)
∀ a, H. a ∈ st ∧ adP0(a, st[a])(H) → adP(a, st[a])(H) ⊦ ((adP0 ⊗_ st))(H) → ((adP ⊗_ st))(H)
Lemma
Sep-forall