 Solution

Here $\mathcal{B}\subseteq \mathcal{A}$ means a substructure, so the smaller carrier contains all named constants, is closed under every function symbol, and has each relation restricted from the larger first-order structure. First-order term evaluation on tuples from $B$ agrees in the two first-order structures. Consequently atomic formulas agree, and induction on Boolean connectives makes every quantifier-free formula agree as well.

Write an arbitrary universal sentence of $T$ as $\forall \vec{x}\theta (\vec{x})$ with quantifier-free matrix. Every tuple from $B$ is also a tuple from $A$, so $\theta$ holds at that tuple in $\mathcal{A}$. Quantifier-free agreement transfers it to $\mathcal{B}$. Since the tuple was arbitrary, $\mathcal{B}$ satisfies the universal sentence. ThusAn arbitrary subset without the substructure operations and relation restrictions would not be the hypothesis used by this formal proof.