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Proof step 6 from proof owns-none-one (
previous step
)
⊦ owns-one(a, r, rs, g) ↔ r ≠ null ∧ r ∈ rs ∧ g(r) = a ∧ (∀ r0. r0 ∈ rs ∧ r0 ≠ r → g(r0) ≠ a)
Lemma
owns-one