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Proof step 3 from proof Sep-eq (previous step)

∀ a. a ∈ s → aP(a) = aP0(a) ⊦ aP ⊗_ s = aP0 ⊗_ s

User applied rule expand right

And obtained the following new goals

Proof step 4:
∀ a. a ∈ s → aP(a) = aP0(a) ⊦ ((aP ⊗_ s))(H) ↔ ((aP0 ⊗_ s))(H)

Rule argument

pos: right 1

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