Equational theories in commutative idempotent monoids It might well be that my question is trivial but I'm not a mathematician, I just need a formalization for algebraic semiotics.
If I have a commutative idempotent monoid, I can define a partial ordering $\preccurlyeq$ as follows: $$a \preccurlyeq b \equiv_{df} a \cdot b = b$$
Now let $E$ be a set of equational axioms (identities) $\{a_1\approx b_1,\dots,a_n\approx b_n\}$ (over elements $a_i,b_i$ of the monoid). I define $\doteq_E$ as the least relation such that


*

*$a\approx b\in E$ implies $a\doteq_E b$,

*the relation $\doteq_E$ is reflexive, symmetric and transitive,

*$a\doteq_E b$ implies $a\cdot c \doteq_E b\cdot c$ for all $a,b,c$.


Is this relation a congruence?
 A: The answer to your question is yes, but let me add a few comments about terminology and notation. 
First, your question makes sense in any commutative monoid and even in any monoid if you change (3) to its two-sided version. Next, it is common practice to write the product in a monoid without the middle dot (that is $ab$ instead of $a \cdot b$). The terms equational axioms and identities have a special meaning in mathematics and are not appropriate in the context of your question. You also need a short argument to show that there is a least relation satisfying your conditions. That being said, one can reformulate your question as follows:

Let $M$ be a monoid and let $R$ be a symmetric relation on $M$. How to define the 
  congruence $\sim_R$ generated by $R$? 

There are two possible answers to this question.  
The first answer is nonconstructive. The set of all congruences containing $R$ is nonempty (it contains the universal relation) and it is closed under intersection. Define the congruence generated by $R$ as the intersection of all congruences containing $R$. 
The second definition is more constructive. Recall that the transitive closure $R^+$ of $R$ is defined as follows: $x\ R^+\ y$ if there is a finite chain $x_0, ..., x_n$ such that $x_0 = x$, $x_n = y$ and $x_i\ R\ x_{i+1}$ for $0 \leqslant i \leqslant n-1$. The 
reflexive transitive closure $R^*$ of $R$ is the union of $R^+$ and the equality relation.
Define the congruence generated by $R$ as $S^*$, where $S = \{(xay, xby) \mid (a,b) \in R, x, y \in M \}$.
You gave essentially a reformulation of this definition: the congruence generated by $R$ is the least relation $\sim_R$ such that
(1) $\sim_R$ contains $R$,
(2) $\sim_R$ is an equivalence relation,
(3) $\sim_R$ is stable under product (in the two-sided case, $a \sim_R b$ implies $ac \sim_R bc$ and $ca \sim_R cb$).
The equivalence between these definitions requires a formal proof, but this answer is already long enough.
