Lawvere algebraic theories as presentation-invariant We can read in a lot of papers, included Lawvere's PhD thesis, that algebraic theories are "an invariant notion of which the usual formalism with operations and equations may be regarded as a ‘presentation’". 
But in what sense an algebraic theory is such an invariant? And invariant for what kind of "transformations"? How can we prove that there is only one algebraic theory for a given type of algebraic structures, e. g. for Boolean rings?
Moreover, it is often said that algebraic theories are "language-free", whereas clones are built from présentations. I'm not sure to understand this claim.
Many thanks, in advance!
 A: As Zhen Lin says in the comments, it's in the same sense that a group is an invariant notion of which a collection of generators and relations is a presentation. The example you give your question is a good one: Boolean rings and Boolean algebras are two different presentations (involving "generators" given by operations and "relations" given by identities that must hold between those operations) of the same Lawvere theory. A short description of this Lawvere theory which does not involve generators and relations is that it is the category of finite sets of cardinality a power of $2$.
In general, if you suspect you have some category $C$ that is the category of models of a Lawvere theory and you want to recover the Lawvere theory itself, you first need the additional data of the forgetful functor $C \to \text{Set}$. Once you have that data, the Lawvere theory, if it exists, is necessarily the "endomorphism Lawvere theory" of this functor, and in particular it is uniquely determined by the structure of $C$ as a concrete category. See this blog post for elaboration and examples. 
