Class of algebras defined by quasi-identities is not closed under taking the quotients

Question

Let $\mathbf{A}$ be an algebra (in the sense of universal algebra) of some signature $\Sigma$. By quasi-identity I mean the formula of the form

$$(\forall x_1) (\forall x_2) \dots (\forall x_n) \left(\left[\bigwedge_{i=1}^{k}t_i(x_1, \dots, x_n)=s_i(x_1, \dots, x_n)\right]\rightarrow t(x_1, \dots, x_n)=s(x_1, \dots, x_n) \right) \;, $$

where $t_i(x_1, \dots, x_n), s_i(x_1, \dots, x_n), t(x_1, \dots, x_n), s(x_1, \dots, x_n)$ are terms (using the algebra operations) with all its variables among $x_1, \dots, x_n$.

Since the class of all algebras (of the considered signature) satisfying some quasi-identity is (allegedly) not a variety in general and clearly such class is closed under taking subalgebras and products, it follows that it is not closed under taking quotients in general.

So the question is:

Is there some (possibly elementary) example of an algebra satisfying some quasi-identity and its quotient where the quasi-identity does not hold?

My only idea was to show this for the cancellation law in some monoid, but I do not see any example (mainly because I do not see how the quotients look like).

Thanks in advance for any help.

Answer 1

Your idea was correct. Take the cancellation law in a commutative monoid. It is satisfied by $\mathbb N$ but not by its quotient $(\{0, 1\}, +)$ (obtained by identifying all positive numbers to $1$). In this latter monoid, you have $0 + 1 = 1 = 1 + 1$, but $0 \not= 1$.

Answer 2

The quasi-identity $\forall x(x+x=0\to x=0)$ holds in $\mathbb Z$ but not in $\mathbb Z/2\mathbb Z$.