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Proof step 9 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 4:
(adP ⊗_ st)(H), ∀ a. a ∈ st → adP(a, st[a]) = adP0(a, st[a]) ⊦ (adP0 ⊗_ st)(H)

Rule argument

[CASEDARG : pos: right 1, ()]

Additional information

[[CONCRETESIMPRULES : ()]]