Is there an algebra $A$ (for example a group) such that $Th(A)$ is logically equivalent to $id(A)$? In other words, is there an algebra $A$ such that
$$
Mod(Th(A))=Var(A)?
$$
Clearly finite algebras do not have this property. It seems that such an algebra should be relatively free.


This question is related to my previous two questions 


1. [Relatively free algebras in a variety generated by a single algebra](https://mathoverflow.net/questions/153743/relatively-free-algebras-in-a-variety-generated-by-a-single-algebra)
2. [relatively free groups in $Var(S\_3)$](https://mathoverflow.net/questions/153883/relatively-free-groups-in-vars-3)


Edition: Only trivial algebra has this property by the comment of Gerhard Paseman. So I ask again the question by $\pm id(A)$ instead of $id(A)$. Is there any algebra A (especially a group) such that $Th(A)$ is logically equivalent to $\pm id(A)$? Here $\pm id(A)$ means the set of identities and negated identities. 


P.S. By negated identity I mean a sentence of the form
$$
\forall x\_1\ldots \forall x\_n: p(x\_1, \ldots, x\_n)\neq q(x\_1, \ldots, x\_n),
$$
where $p$ and $q$ are terms. Is there any negated identity in a non-trivial algebra? Clearly there are no negated identities in groups but if we add constants to the language of groups there will be many negated identities.



The existence of an algebra whose set of identities and first order theory are equivalent
I imagine that definitions of $Mod, Th, Var$ and so on have not changed since I saw them decades ago. The trivial one element algebra in any finite type (and likely any infinite type) is an easy example which satisfies $Mod(Th(\textbf{A})) \approx Var(\textbf{A})$. Since it is expected that $Th(\textbf{A})$ is strong enough to indicate whether $\textbf{A}$ has more than one element, this is the only example to be expected (Thank you Joel).


Gerhard "Ask Me About Trivial Algebra" Paseman, 2014.01.09


