On the definition of an Interpretation Isomorphism

Question

I wish to check my understanding of an interpretation isomorphism, as defined (although paraphrased) in Boolos' Computability and Logic:

Two interpretations $P$ and $Q$ are isomorphic iff there is a map $j:P\to Q$ such that

1. $j\left(f^P(p_1,\ldots,p_n)\right) = f^Q\left(j(p_1),\ldots,j(p_n)\right).$
2. $R^P(p_1,\ldots,p_n)$ if and only if $R^Q\left(j(p_1),\ldots,j(p_n)\right)$.
3. $j\left(c^P\right) = c^Q$.

I was wondering why the first two conditions act directly on members of $|P|$ and $|Q|$, as opposed to acting on closed terms. In the latter case, conditions $(1)$ and $(2)$ would be

1. $j\left(f^P(t_1^P,\ldots,t_n^P)\right) = f^Q\left(j(t_1^Q),\ldots,j(t_n^Q)\right)$.
2. $R^P(t_1^P,\ldots,t_n^P)$ if and only if $R^Q\left(j(t_1^P),\ldots,j(t_n^P)\right)$.

for any closed terms $t_i$.

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Is the following the reason for choosing the former definition?

If we use the latter definition, the Isomorphism Lemma would fail.

(Isomorphism Lemma) If there is an isomorphism between two interpretations $P$ and $Q$ of the same language $L$, then for every sentence $A$ of $L$ we have $$P\models A \ \ \ \ \text{ if and only if } Q\models A.$$

As to how it could fail, consider $$|P| = \mathbb{N} = \{0,1,2,3,...\}$$ with the only predicate $>$, and $$|Q| = -\mathbb{N} = \{0,-1,-2,-3,...\}$$ with the only predicate $>$ (yes, both $P$ and $Q$ with the same predicate), where the isomorphism, as per the second definition, is defined as $$n\mapsto -n \ \ \ \ \text{and} \ \ \ \ >\ \mapsto \ >.$$

Suppose we have only one constant $c_0$ in our language, and that $c_0^P=0$. As we are using the objectual approach, we have that $$P \models \exists x(x>0)$$ as we can append a new constant $c_1$ to our language and map it to $1$. However, such sentence $\exists x(x>0)$ is not true in $Q$, as no element of $|Q|$ is greater than $0$.

Answer 1

There is much more to an interpretation than its closed terms. So a definition of isomorphism that only mentions the elements named by closed terms would be very strange indeed.

For example, a group is an interpretation of the language of groups, but the only element of a group named by a closed term is the identity element. So by your definition, any two groups would be isomorphic as interpretations. This is obviously undesirable.