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Proof step 12 from proof Sep-eq (previous step)

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)

User applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: a abort: F discard: F computed: T

Additional information

Used terms: a