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Proof step 11 from proof Sep-eq (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
adP0(a, st[a])(H0), a ∈ st, ((adP0 ⊗_ st))(H), ∀ a. a ∈ st → adP(a, st[a]) = adP0(a, st[a]) ⊦ adP(a, st[a])(H0), ((adP ⊗_ st))(H)

Proof step 11 from proof Sep-eq (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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