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Proof step 10 from proof Sep-eq (previous step)

∀ a. a ∈ st → adP(a, st[a]) = adP0(a, st[a]) ⊦ adP ⊗_ st = adP0 ⊗_ st

User applied rule expand right

And obtained the following new goals

Proof step 9:
∀ a. a ∈ st → adP(a, st[a]) = adP0(a, st[a]) ⊦ (adP ⊗_ st)(H) ↔ (adP0 ⊗_ st)(H)

Rule argument

[fmapos-arg: pos: right 1]

Additional information

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