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Proof step 1 from proof Sep-any (conclusion)

⊦ st ≠ ∅ → (((adP ⊗_ st * P))(H) ↔ (∃ a, d, st0. ¬ a ∈ st0 ∧ st = st0[a, d] ∧ ((adP(a, d) * adP ⊗_ st0 * P))(H)))

System applied rule normalize

And obtained the following new goals

Rule argument

Additional information