Given a set $X$ and a class $K$ of universal algebras of the same type $F$, an $F$-algebra $U$, generated by $X$, is said to be free over $X$ for the class $K$ if, for every $F$-algebra $A$ in $K$ and every function $\alpha\colon X\to A$ there exists exactly one $F$-morphism $\hat{\alpha}\colon U\to A$ extending $\alpha$. The problem here is that, in general, the free $F$-algebra over $X$ is not a member of $K$. To solve this problem, and conclude that, for instance, all varieties $V$ of algebras contain the $V$-free algebra over $X$, Burris-Sankappanavar in their Universal Algebra textbook at page 73, introduce this construction:
Take the term algebra $T(X)$ of $F$-terms over a set $X$, take the smallest congruence $\theta$ on $T(X)$ to which corresponds a quotient $T(X)/\theta\in IS(K)$ (the class of algebras isomorphic to a subalgebra of a member of K), then $T(X)/\theta$ is in $K$ when, for instance, $K$ is a variety (a class of algebras closed under subalgebras, homomorphic images and direct products of its members). More precisely, let $$\theta_K(X):=\bigcap\Phi_K(X)$$ and $$\Phi_K(X):=\{\theta\in Cong(T(X)): T(X)/\theta\in IS(K)\}$$ Then define the $K$-free algebra over $X$ as: $$F_K(X):=T(X)/\theta_K(X)$$
My question is: how do we know, in general, that $\Phi_K(X)$ is non-empty?
Actually it could be empty. This is the case for example if $K$ is empty, because in that case $IS(K)$ is empty, so there is no congruence such that $T(X)/\theta\in IS(K)$.
But if you look at the remarks after Definition 10.9, it says :
In fact, $\theta_K(X)$ should be defined as the infimum (or meet) of the set $\Phi_K(X)$ in the lattice of congruences of $T(X)$. If $\Phi_K(X)$ is non-empty, then this infimum is the intersection of all the congruences in $\Phi_K(X)$, but if is empty then its infimum is simply the largest congruence $\nabla$.