[proofplan] We first identify every $\mathcal L_{\mathrm{ring}}(K)$-term in the variables $x_1,\dots,x_n$ with a polynomial in $K[X_1,\dots,X_n]$ having the same value at every point of $K^n$. Therefore each atomic formula defines the zero set of a polynomial, hence a Zariski closed subset of $K^n$. Finally, we use structural induction on the quantifier-free formula: the Boolean connectives correspond exactly to complement, finite intersection, and finite union, and the class of constructible subsets is closed under those operations. [/proofplan]

[step:Replace every ring term by its polynomial function]Let $R := K[X_1,\dots,X_n]$. For every $\mathcal L_{\mathrm{ring}}(K)$-term $\tau(x_1,\dots,x_n)$, define a polynomial $P_\tau \in R$ by induction on the construction of $\tau$ as follows. If $\tau$ is the variable $x_i$, set $P_\tau := X_i$. If $\tau$ is the constant symbol for $c \in K$, set $P_\tau := c \in R$. If $\tau$ is built from terms $\sigma$ and $\rho$, set \begin{align*} P_{\sigma+\rho} &:= P_\sigma + P_\rho,\\ P_{\sigma-\rho} &:= P_\sigma - P_\rho,\\ P_{\sigma\cdot \rho} &:= P_\sigma P_\rho. \end{align*} If the chosen presentation of the ring language has unary negation rather than binary subtraction, set $P_{-\sigma}:=-P_\sigma$. We prove by induction on terms that for every $a=(a_1,\dots,a_n)\in K^n$, \begin{align*} \tau^K(a)=P_\tau(a), \end{align*} where $\tau^K:K^n\to K$ denotes the function obtained by interpreting the term $\tau$ in the field $K$. The assertion is immediate for variables and constant symbols by the definitions of $X_i$ and of constant polynomials. The induction steps follow because the interpretations of $+$, $-$, and $\cdot$ in $K$ agree with addition, subtraction, and multiplication of polynomial values.[/step]

[guided]The point of this step is to remove the model-theoretic syntax from the terms. We define a map from terms to polynomials by following the grammar of the language. A variable $x_i$ becomes the coordinate polynomial $X_i$, and a parameter $c \in K$ becomes the constant polynomial $c$. The function symbols of the ring language are then interpreted by the corresponding polynomial operations: \begin{align*} P_{\sigma+\rho} &:= P_\sigma + P_\rho,\\ P_{\sigma-\rho} &:= P_\sigma - P_\rho,\\ P_{\sigma\cdot \rho} &:= P_\sigma P_\rho. \end{align*} If negation is primitive in the chosen language, we also set $P_{-\sigma}:=-P_\sigma$. We now verify that this syntactic replacement preserves values. For every term $\tau(x_1,\dots,x_n)$, let \begin{align*} \tau^K:K^n &\to K \end{align*} be the function obtained by interpreting $\tau$ in the field $K$. We prove, by induction on the construction of $\tau$, that \begin{align*} \tau^K(a)=P_\tau(a) \end{align*} for every point $a=(a_1,\dots,a_n)\in K^n$. For $\tau=x_i$, both sides are $a_i$. For a constant symbol $c\in K$, both sides are $c$. If the identity is known for $\sigma$ and $\rho$, then the interpretation of the term $\sigma+\rho$ satisfies \begin{align*} (\sigma+\rho)^K(a)=\sigma^K(a)+\rho^K(a)=P_\sigma(a)+P_\rho(a)=(P_\sigma+P_\rho)(a)=P_{\sigma+\rho}(a). \end{align*} The subtraction and multiplication cases are identical computations using the field operations in $K$. Thus every term in the language, even with parameters from $K$, evaluates as a polynomial function on $K^n$.[/guided]

[step:Show that atomic formulas define Zariski closed subsets] Let $\psi(x_1,\dots,x_n)$ be an atomic $\mathcal L_{\mathrm{ring}}(K)$-formula. Since the only relation symbol in the language of rings is equality, $\psi$ has the form \begin{align*} \sigma(x_1,\dots,x_n)=\rho(x_1,\dots,x_n) \end{align*} for two $\mathcal L_{\mathrm{ring}}(K)$-terms $\sigma$ and $\rho$. Let $P_\sigma,P_\rho\in K[X_1,\dots,X_n]$ be the polynomials attached to these terms in the previous step, and define \begin{align*} h_{\psi}:=P_\sigma-P_\rho \in K[X_1,\dots,X_n]. \end{align*} For every $a\in K^n$, \begin{align*} K\models \psi(a) &\iff \sigma^K(a)=\rho^K(a)\\ &\iff P_\sigma(a)=P_\rho(a)\\ &\iff h_{\psi}(a)=0. \end{align*} Therefore \begin{align*} \psi(K)=Z(h_{\psi}), \end{align*} so $\psi(K)$ is Zariski closed in $K^n$, and hence constructible. [/step]

[step:Close the argument under Boolean connectives]Let $\mathcal C_n$ denote the Boolean algebra of subsets of $K^n$ generated by all polynomial zero sets $Z(f)$ with $f\in K[X_1,\dots,X_n]$. By definition, the elements of $\mathcal C_n$ are precisely the constructible subsets of $K^n$ in the sense used in the statement, and $\mathcal C_n$ is closed under finite unions, finite intersections, and complements in $K^n$. We prove by structural induction on quantifier-free formulas $\theta(x_1,\dots,x_n)$ that \begin{align*} \theta(K):=\{a\in K^n:K\models \theta(a)\} \end{align*} belongs to $\mathcal C_n$. The atomic case was proved in the previous step. If $\theta=\neg \alpha$, then \begin{align*} \theta(K)=K^n\setminus \alpha(K), \end{align*} which belongs to $\mathcal C_n$ whenever $\alpha(K)\in\mathcal C_n$. If $\theta=\alpha\wedge\beta$, then \begin{align*} \theta(K)=\alpha(K)\cap \beta(K), \end{align*} which belongs to $\mathcal C_n$ whenever $\alpha(K),\beta(K)\in\mathcal C_n$. If $\theta=\alpha\vee\beta$, then \begin{align*} \theta(K)=\alpha(K)\cup \beta(K), \end{align*} which belongs to $\mathcal C_n$ whenever $\alpha(K),\beta(K)\in\mathcal C_n$. Any additional Boolean connective, such as implication or biconditional, is an abbreviation in terms of $\neg$, $\wedge$, and $\vee$, so the same conclusion follows for all quantifier-free formulas. Applying this induction to the given formula $\varphi(x_1,\dots,x_n)$ gives \begin{align*} \varphi(K)\in \mathcal C_n. \end{align*} Thus $\varphi(K)$ is constructible in $K^n$.[/step]

[guided]We now pass from atomic formulas to arbitrary quantifier-free formulas. Define $\mathcal C_n$ to be the Boolean algebra of subsets of $K^n$ generated by the polynomial zero sets \begin{align*} Z(f)=\{a\in K^n:f(a)=0\}, \qquad f\in K[X_1,\dots,X_n]. \end{align*} Thus a subset of $K^n$ is constructible exactly when it lies in $\mathcal C_n$. The reason for introducing $\mathcal C_n$ is that quantifier-free formulas are built from atomic formulas using only Boolean operations, and $\mathcal C_n$ is designed to be closed under those same operations. We prove the desired statement by structural induction on a quantifier-free formula $\theta(x_1,\dots,x_n)$. The base case is when $\theta$ is atomic. In that case, the previous step produced a polynomial $h_\theta\in K[X_1,\dots,X_n]$ such that \begin{align*} \theta(K)=Z(h_\theta), \end{align*} so $\theta(K)\in\mathcal C_n$. For the induction step, suppose first that $\theta=\neg\alpha$ and assume $\alpha(K)\in\mathcal C_n$. A point $a\in K^n$ satisfies $\neg\alpha$ in $K$ exactly when it does not satisfy $\alpha$, hence \begin{align*} \theta(K)=K^n\setminus \alpha(K). \end{align*} Since $\mathcal C_n$ is a Boolean algebra, it is closed under complements in $K^n$, so $\theta(K)\in\mathcal C_n$. Next suppose $\theta=\alpha\wedge\beta$ and assume $\alpha(K),\beta(K)\in\mathcal C_n$. A point $a\in K^n$ satisfies $\alpha\wedge\beta$ exactly when it satisfies both $\alpha$ and $\beta$, so \begin{align*} \theta(K)=\alpha(K)\cap \beta(K). \end{align*} Closure of $\mathcal C_n$ under finite intersections gives $\theta(K)\in\mathcal C_n$. Finally suppose $\theta=\alpha\vee\beta$ and assume $\alpha(K),\beta(K)\in\mathcal C_n$. A point $a\in K^n$ satisfies $\alpha\vee\beta$ exactly when it satisfies at least one of $\alpha$ and $\beta$, so \begin{align*} \theta(K)=\alpha(K)\cup \beta(K). \end{align*} Closure of $\mathcal C_n$ under finite unions gives $\theta(K)\in\mathcal C_n$. These cases cover the usual formation rules for quantifier-free formulas. If the syntax also permits implication or biconditional as primitive-looking connectives, they are definitional abbreviations: \begin{align*} \alpha\implies\beta &\equiv \neg\alpha\vee\beta,\\ \alpha\leftrightarrow\beta &\equiv (\alpha\wedge\beta)\vee(\neg\alpha\wedge\neg\beta). \end{align*} Therefore they introduce no new set-theoretic operations beyond complement, finite intersection, and finite union. Applying the structural induction to the given formula $\varphi$ yields \begin{align*} \varphi(K)\in\mathcal C_n, \end{align*} which is exactly the assertion that $\varphi(K)$ is constructible in $K^n$.[/guided]