2 minor correction

There is a major difference between elementary equivalence and elementary embedding. Moreover, in this case, the actual embedding is somewhat ambiguous. First, let me recap some often confused terminology.

Two models are elementary equivalent if they satisfy the same first-order sentences. Any two isomorphic models are always elementary equivalent. An elementary embedding is a map j:A→B such that, for all first-order formulas φ(v1,...,vk) and all a1,...,ak ∈ A, A ⊧ φ(a1,...,ak) iff B ⊧ φ(j(a1),...,j(ak)). An isomorphism is always an elementary embedding.

The notation A ≺ B means that A is an elementary submodel of B, i.e. the inclusion map A ⊆ B is an elementary embedding from A into B. In your context, the isomorphism (or its inverse) is not the proposed elementary embedding, it is the inclusion map which is in question: it is elementary for Σn formulas, but not elementary for all first-order formulas.

1

There is a major difference between elementary equivalence and elementary embedding. Moreover, in this case, the actual embedding is. First, let me recap some often confused terminology.

Two models are elementary equivalent if they satisfy the same first-order sentences. Any two isomorphic models are always elementary equivalent. An elementary embedding is a map j:A→B such that, for all first-order formulas φ(v1,...,vk) and all a1,...,ak ∈ A, A ⊧ φ(a1,...,ak) iff B ⊧ φ(j(a1),...,j(ak)). An isomorphism is always an elementary embedding.

The notation A ≺ B means that A is an elementary submodel of B, i.e. the inclusion map A ⊆ B is an elementary embedding from A into B. In your context, the isomorphism (or its inverse) is not the proposed elementary embedding, it is the inclusion map which is in question: it is elementary for Σn formulas, but not elementary for all first-order formulas.