= 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>. Thus
$$
\boxed{\mathcal A\models T,\ \mathcal B\subseteq\mathcal A
\Longrightarrow\mathcal B\models T.}
$$
An arbitrary <subset> without the <substructure> operations and relation restrictions would not be the hypothesis used by this <formal proof>.
Back to article page
