Formal group operation This exercise appears in An Invitation to General Algebra and Universal Constructions by George M. Bergman.

By "the set of all terms in the elements of $X$ under the formal group operations $\mu, \iota, e$" we shall mean a set $T$ which is
  
  
*
  
*given with functions
  $\mathrm{symb}_T : X \to T, \mu_T : T^2 \to T, \iota_T : T \to T,$ and $e_T : T^0 \to T$,
  such that 
  
*each of these maps is one-to-one, their images are disjoint, and $T$ is the union of those images, and
  
*$T$ is generated by $\mathrm{symb}_T(X)$ under the operations $\mu_T, \iota_T,$ and $e_T$; i.e., it has no proper subset which contains $\mathrm{symb}_T(X)$ and is closed under those operations. 
  

Now, the exercise asks whether the third condition follows from the first two. I am inclined to believe that it does not, but I am not sure how to proceed with a proof. 
A question which may or may not be relevant is also whether these axioms forbid a term from being the image of a term, which is also the image of another term..., continuing like this and never reaching $\mathrm{symb}_T(x)$ for some $x \in X$. I am used to bottom-up definitions where terms are given by induction on "depth" or "length", but I can't get my head around this presentation.
Thanks.
 A: Terms can be visualized via "term trees".
The term trees in the language of groups over the set $X$ can be described
as finite, rooted, labeled trees satisfying the following conditions.
Each node has degree at most $2$. (Degree = number of children.)
A node of degree $2$ is labeled with the symbol $\mu$
(denoting multiplication).
A node of degree $1$ is labeled with the symbol $\iota$ (inverse).
A node of degree $0$ is labeled with an element of $X$
(generator) or with
the symbol $e$ (identity).
Let $T$ be the set of these labeled trees. Define $\textrm{symb}_T$
so that it
maps $x\in X$ to the 1-node tree with label $x$. Define
$e_T$ so that it
maps the single element of $T^0$ to the 1-node tree with label $e$.
Define $\mu_T$ so that it
maps the pair $(s,t)\in T\times T$ to the
tree in $T$ with root of degree $2$ labeled $\mu$,
whose left child is the root of a subtree equal to
$s$, and whose right
child is the root of a subtree equal to
$t$.
Define $\iota_T$ so that it maps $s\in T$ to the
tree with root of degree $1$ labeled $\iota$,
whose child is the root of a subtree equal to $s$.
It is not hard to show that all three of Bergman's
conditions hold for the data $T$, $\mu_T$, $\iota_T$, $e_T$.
Now to address the problem: "Does
the third of Bergman's conditions follow from the first two?"
Answer: No.
Let $T'$ be a set of possibly infinite, rooted, labeled trees
defined by the same conditions. To make sure this is a set, restrict to those possibly infinite trees where every node
has a finite path to the root. Conditions 1 and 2 still
hold for $T'$, but Condition 3 does not. The subset of $T'$ generated
by $\textrm{symb}_T(X)$ under the operations is just the subset of finite labeled trees.
