How do we prove that an interpretation A is isomorphic to itself? Prove that an A is isomorphic to A, where A is an interpretation.
So far I know that there is a correspondence from A to A via the identity function because $id(x) = x$ for every $x$. This proves condition $I_2$.
But how do I go about proving the rest of the conditions?
Interpretations $P$ and $Q$ are isomorphic if and only if there is a correspondence $j$ between the two and if the following conditions hold:
$I_1) \, \, R^P(p_1,...,p_n)$ if and only if $R^Q(j(p_1),...,j(p_n))$
$I_2) \, \, j(c^P) = c^Q$
$I_3) \, \, j(f^P(p_1,...,p_n)) = f^Q(j(p_1),...,j(p_n))$.
 A: The model theory bible by Chang and Keisler from 1973 defines two models $\mathfrak{A}$ and $\mathfrak{A}'$, or in your words interpretations, to be isomorphic iff there is a bijection $f$ mapping $A$ onto $A'$ (where $A$ and $A'$ are the domains of resp. $\mathfrak{A}$ and $\mathfrak{A}'$), satisfying the following conditions:


*

*for each $n$-placed relation R of $\mathfrak{A}$ and the corresponding relation $R'$ of $\mathfrak{A}$: $R(a_1,.,..,a_n)$ iff $R'(f(a_1),..,f(a_n)$;

*for each $m$-placed function $G$ on $\mathfrak{A}$ and the corresponding function $G'$ of $\mathfrak{A}'$: $f(G(a_1,...,a_m)) = G'(f(a_1),...,f(a_n))$;

*for each constant $c$ of $\mathfrak{A}$ and the corresponding constant $c'$ of $\mathfrak{A}'$: $f(c) = c'$.


This definition is clearly equivalent to your definition. Now define a function $f: A \rightarrow A$, s.t. for every $a_i \in A$, $a_i \mapsto a_i$. Clearly $f$ is then the identity function, which is a bijective function. So we just need to check the three conditions.


*

*Let $R(a_1,...,a_n)$ be some $n$-placed relation $R$ of $\mathfrak{A}$. We have $a_i = f(a_i)$ for all $a_i \in R$, hence $R(a_1,...,a_n) = R(f(a_1),...,f(a_n))$, so we have $R(a_1,...,a_n)$ iff $R(f(a_1),...,f(a_n))$.

*Let $G(a_1,...,a_m)$ be some $m$-placed function $G$ on $\mathfrak{A}$. We have $a_i = f(a_i)$ for all elements of $A$, so also $a_i = f(a_i)$ for all arguments $a_i$ of $G$, hence $G(a_1,...,a_m) = G(f(a_1),...,f(a_n))$. Now $G(a_1,...,a_m)=a_j$ for some element $a_j \in A$, so $G(a_1,...,a_m) = G(f(a_1),...,f(a_n))=a_j=f(a_j)$. From this we can conclude that $f(G(a_1,...,a_m)) = f(a_j) = G(f(a_1),...,f(a_n))$.

*Let $c$ be some constant of $\mathfrak{A}$, that means that some constant $c^\mathcal{L}$ of our language $\mathcal{L}$ is interpreted by an element $a_i \in A$, which we denote by $c$. We have $a_i = f(a_i)$ for all elements of $A$, hence we have $c = f(c)$.


From the foregoing we can conclude that $\mathfrak{A}$ is isomorphic to $\mathfrak{A}$, since there is a bijection $f: A \rightarrow A$ satisfying all the conditions above.
