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Proof step 7 from proof Sep-eq (
previous step
)
aP(a)(H0), a ∈ s, ∀ a. a ∈ s → aP(a) = aP0(a) ⊦ # s = 0, ((aP ⊗_ s))(H) ↔ ((aP0 ⊗_ s))(H), s = ∅, aP0(a)(H0)
User applied rule all left
And closed the goal!
Rule argument
pos: left 3 terms: a abort: F discard: F computed: T
Additional information
Used terms: a