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Proof step 14 from proof abs-idset-fwd (
previous step
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⊦ id(apebs0, apebs1, ns) ↔ # apebs0 = # apebs1 ∧ (∀ n. n ∈ ns ∧ n < # apebs0 ∧ n < # apebs1 → apebs0[n] = apebs1[n])
A spec-lemma