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Proof step 4 from proof Sep-eq (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 5:
((adP ⊗_ st))(H), ∀ a. a ∈ st → adP(a, st[a]) = adP0(a, st[a]) ⊦ ((adP0 ⊗_ st))(H)
Proof step 9:
((adP0 ⊗_ st))(H), ∀ a. a ∈ st → adP(a, st[a]) = adP0(a, st[a]) ⊦ ((adP ⊗_ st))(H)

Rule argument

pos: right 1

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