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

aP0(a)(H0), a ∈ s, ((aP0 ⊗_ s))(H), ∀ a. a ∈ s → aP(a) = aP0(a) ⊦ # s = 0, s = ∅, aP(a)(H0), ((aP ⊗_ s))(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