 Solution

The set of closed-term values is nonempty because the first-order language has a constant. It contains the value of every constant. If $f$ is an $r$-ary function symbol and $t_1,\dots ,t_r$ are closed terms, thenwhich is another closed-term value. Hence the set is closed under every operation. Restrict the operations and relations of $\mathcal{A}$ to it; this defines a substructure $\mathcal{B}$ whose carrier is exactly the specified set.

Part (i) now applies to the universal first-order theory, givingIt is the substructure generated by the first-order language's constants, often called the ground-term substructure. The source's constant-symbol assumption ensures its nonemptiness, even when no other elements can be generated.