Finite algebras can be given by finite presentations Let $A$ be a finite algebra in variety $V$ with finite language $L$.  Then let $X$ be the basis of $A$, which is obviously finite since $A$ is finite.  I know that a finite presentation $\langle X|R \rangle$ is one where both $X$ and $R$ are finite so I just need to show $R$ is finite.
Let $I$ be the set of identities of $V$, $I = \{t_a(u_1,...u_{n_a}) = s_a(u_1,...,u_{n_a})\}$.  Then $R = \{t_a(u_1,...u_{n_a}) = s_a(u_1,...,u_{n_a}) | a \in I, u_1,..., u_{n_a} \in T\}$ where $T$ is a term algebra.  I don't get what a term algebra is, but I think that it can be shown that the set of identities is finite so $R \subset I$ is finite.  Is that correct?  What else do I need to show?
 A: I don't quite understand the approach you describe, but since there are (probably) infinitely many terms in the language (remember, we can compose functions), I don't see why it should give a finite set of relations.
However, there's a much simpler argument: basically, we look at the "multiplication table" for our structure - that is, our relations are the "simple" equations which describe how all the function symbols operate.

First, let's assume $X=A$. Say that a basic equation is one of the form $$f(c_1, ..., c_n)=d$$ where $c_1, ..., c_n, d$ are elements of $A$ and $f$ is an $n$-ary function symbol in our language. The set of basic equations is finite since both $A$ and our language are finite, so taking $R$ to be the set of basic equations  which are true in $A$ we get a finite set of relations as desired (check that $A=\langle A\vert R\rangle$).

Now what about when $X\subsetneq A$? Well, we'll just repeat the above idea. For $a\in A$, let $(t_a(\overline{x}), \overline{c}_a)$ be a pair where $t$ is a term in the displayed variables, $\overline{c}_a$ is a tuple of elements of $X$ with the same arity as $\overline{x}$, and in $A$ we have $t_a(\overline{c})=a$. Say that a $X$-basic equation is an equation of the form $$f(t_{c_1}, ..., t_{c_n})=t_{d}$$ for $c_1, ..., c_n, d\in A$ and $f$ an $n$-ary function symbol in our language. Again, if we let $R$ be the set of $X$-basic equations holding in $A$, we have $A=\langle X\vert R\rangle$.
