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Proof step 4 from proof idset-idexpt (previous step)

⊦ id○(apebs, apebs0, n) ↔ # apebs = # apebs0 ∧ (∀ m. m < # apebs ∧ m ≠ n → apebs[m] = apebs0[m])

A spec-lemma