'Finitely presented' implies 'always finite presented' for algebraic theories In this MO question it is proven the answer is yes for modules. The proof given relies on the snake lemma, which does not generally make sense in the category of rings, groups, monoids, etc.
It seems strange to me that this implication would hold for modules but not for rings, groups, monoids, lattices etc. Maybe naively, I think it's true for any algebraic theory - if you can present something finitely, then you cannot contrive an infinite presentation. I really have no idea how to prove it though.
So, in which algebraic theories does 'finitely presented' impliy 'always finitely presented'?
In more detail, I am asking for which Lawvere theories do the categories of models satisfy the following property: If there exists a finite presentation of a model, then every presentation on finitely many generators of this model is finite.
 A: I will work in a category $\mathsf{Alg}(\tau)$ of algebraic structures of a given type $\tau$ (in the sense of universal algebra).
Let $A$ be a finitely presented algebra. Choose some surjective homomorphism $\phi : \langle x_1,\dotsc,x_n \rangle \to A$ such that the kernel is a finitely generated congruence. Let $y_1,\dotsc,y_m$ be another generating set of $A$ and consider the surjective homomorphism $\psi : \langle y_1,\dotsc,y_m \rangle \to A$; we want to prove that the kernel of $\psi$ is finitely generated, too.
We define $\alpha : \langle x_1,\dotsc,x_n,y_1,\dotsc,y_m \rangle \to \langle x_1,\dotsc,x_n \rangle$ as follows: We map $x_i \mapsto x_i$, and we map $y_j$ to some preimage of $y_j \in A$ under $\phi : \langle x_1,\dotsc,x_n \rangle \twoheadrightarrow A$. We also obtain $\lambda : \langle x_1,\dotsc,x_n,y_1,\dotsc,y_m \rangle \twoheadrightarrow A$, which is also $\phi \circ \alpha$.
The kernel of $\alpha$ is generated (as a congruence) by the elements $(y_j,\alpha(y_j))$, hence it is finitely generated. The kernel of $\lambda$ is generated by these elements and lifts of the generators of $\ker(\phi)$; hence it is finitely generated. Since the kernel of $\lambda$ surjects onto the kernel of $\psi$, we are done.
(The proof is a bit sketchy. Perhaps I will add some details later.)
