[proofplan] We show that both infinitude and the finite-or-cofinite property are determined by first-order finite patterns over the parameter type $p(z)$. Infinitude transfers because the assertion that $\varphi(x,z)$ has at least $m$ distinct realizations is a first-order condition on $z$. If a definable subset in some elementary extension split $\varphi(N,e')$ into two infinite pieces, then all finite lower-bound versions of that split would form a small partial type together with $p(z)$. Saturation realizes this pattern in $S$, and a second saturation argument moves the parameter from an arbitrary realization of $p$ in $S$ back to the distinguished tuple $e$, contradicting the assumed strong minimality of $\varphi(S,e)$. [/proofplan] [step:Transfer infinitude from the saturated witness to every realization of the parameter type] Let $\mathbb{N}=\{1,2,3,\dots\}$. For any tuple $d$ in a model of $T$, write $\operatorname{tp}(d/\varnothing)$ for the complete first-order type of $d$ over the empty parameter set. Fix a model $M\models T$, a tuple $e'\in M^{|z|}$ realizing $p(z)$, and an elementary extension $N\succeq M$. For each $m\in \mathbb{N}$, define the $L$-formula $\iota_m(z)$ by \begin{align*} \iota_m(z) := \exists x_1\cdots \exists x_m\left(\bigwedge_{i=1}^m \varphi(x_i,z)\wedge \bigwedge_{1\leq i<j\leq m} x_i\neq x_j\right). \end{align*} Since $\varphi(S,e)$ is infinite, $S\models \iota_m(e)$ for every $m\in \mathbb{N}$. Because $e$ realizes the complete type $p(z)$, each formula $\iota_m(z)$ belongs to $p(z)$. Since $e'$ also realizes $p(z)$, $M\models \iota_m(e')$ for every $m\in \mathbb{N}$. As $M\preceq N$, elementarity gives $N\models \iota_m(e')$ for every $m\in \mathbb{N}$. Therefore $\varphi(N,e')$ is infinite. [/step] [step:Convert a failure of strong minimality into finite splitting conditions] Suppose, toward a contradiction, that $\varphi(N,e')$ has a definable subset that is neither finite nor cofinite. Then there are an $L$-formula $\psi(x,y)$ and a tuple $a\in N^{|y|}$ such that, defining \begin{align*} A := \{c\in \varphi(N,e') : N\models \psi(c,a)\}, \end{align*} both $A$ and $\varphi(N,e')\setminus A$ are infinite. For $m,n\in \mathbb{N}$, define the $L$-formula $\theta_{m,n}(z,y)$ by \begin{align*} \theta_{m,n}(z,y) := {}& \exists u_1\cdots \exists u_m \exists v_1\cdots \exists v_n \Bigg( \bigwedge_{i=1}^m \bigl(\varphi(u_i,z)\wedge \psi(u_i,y)\bigr) \\ &\wedge \bigwedge_{j=1}^n \bigl(\varphi(v_j,z)\wedge \neg \psi(v_j,y)\bigr) \wedge \bigwedge_{1\leq i<i'\leq m} u_i\neq u_{i'} \\ &\wedge \bigwedge_{1\leq j<j'\leq n} v_j\neq v_{j'} \wedge \bigwedge_{\substack{1\leq i\leq m\\1\leq j\leq n}} u_i\neq v_j \Bigg). \end{align*} The infinitude of $A$ and of $\varphi(N,e')\setminus A$ gives \begin{align*} N\models \theta_{m,n}(e',a) \end{align*} for every $m,n\in \mathbb{N}$. [guided] Assume that strong minimality fails in the elementary extension $N$. This means that some formula $\psi(x,y)$, with some parameter tuple $a\in N^{|y|}$, defines a subset of $\varphi(N,e')$ that is infinite and whose complement inside $\varphi(N,e')$ is also infinite. We name this subset \begin{align*} A := \{c\in \varphi(N,e') : N\models \psi(c,a)\}. \end{align*} The point is to express this failure using only first-order finite approximations. First-order logic cannot say in one formula that $A$ is infinite, but it can say that $A$ has at least $m$ distinct elements, for each fixed $m\in\mathbb{N}$. Likewise it can say that the complement has at least $n$ distinct elements. For $m,n\in\mathbb{N}$, define $\theta_{m,n}(z,y)$ to assert that there are $m$ distinct realizations of $\varphi(x,z)\wedge \psi(x,y)$ and $n$ distinct realizations of $\varphi(x,z)\wedge \neg\psi(x,y)$, all distinct from each other: \begin{align*} \theta_{m,n}(z,y) := {}& \exists u_1\cdots \exists u_m \exists v_1\cdots \exists v_n \Bigg( \bigwedge_{i=1}^m \bigl(\varphi(u_i,z)\wedge \psi(u_i,y)\bigr) \\ &\wedge \bigwedge_{j=1}^n \bigl(\varphi(v_j,z)\wedge \neg \psi(v_j,y)\bigr) \wedge \bigwedge_{1\leq i<i'\leq m} u_i\neq u_{i'} \\ &\wedge \bigwedge_{1\leq j<j'\leq n} v_j\neq v_{j'} \wedge \bigwedge_{\substack{1\leq i\leq m\\1\leq j\leq n}} u_i\neq v_j \Bigg). \end{align*} The variables $u_i$ name points on the $\psi$-side of the split, and the variables $v_j$ name points on the complementary side. Since $A$ is infinite, we can choose $m$ distinct elements of $A$. Since $\varphi(N,e')\setminus A$ is infinite, we can choose $n$ distinct elements of the complement. These choices witness \begin{align*} N\models \theta_{m,n}(e',a) \end{align*} for every $m,n\in\mathbb{N}$. [/guided] [/step] [step:Realize the finite splitting pattern in the saturated witness model] Let $\Sigma(z,y)$ be the partial type over $\varnothing$ defined by \begin{align*} \Sigma(z,y) := p(z)\cup \{\theta_{m,n}(z,y): m,n\in \mathbb{N}\}. \end{align*} Every finite subset of $\Sigma(z,y)$ is satisfiable in $N$, because $e'$ realizes $p(z)$ and $N\models \theta_{m,n}(e',a)$ for all $m,n\in\mathbb{N}$. To transfer finite satisfiability to $S$, let $\Sigma_0(z,y)\subset \Sigma(z,y)$ be finite, and let $\sigma_0$ be the first-order $L$-sentence asserting that some tuples $z,y$ satisfy all formulas in $\Sigma_0(z,y)$. The previous sentence gives $N\models \sigma_0$. Since $T$ is complete and $S,N\models T$, the sentence $\sigma_0$ is true in every model of $T$ if it is true in one model of $T$; hence $S\models \sigma_0$. Therefore every finite subset of $\Sigma(z,y)$ is satisfiable in $S$. The cardinality of $\Sigma(z,y)$ is at most $|T|+|z|+\aleph_0$, hence is less than $\kappa$ by the hypothesis $\kappa>|T|+|z|$. Since $S$ is $\kappa$-saturated, there are tuples $e_0\in S^{|z|}$ and $b_0\in S^{|y|}$ such that \begin{align*} S\models \Sigma(e_0,b_0). \end{align*} In particular, $e_0$ realizes $p(z)$ and \begin{align*} S\models \theta_{m,n}(e_0,b_0) \end{align*} for every $m,n\in\mathbb{N}$. [/step] [step:Move the realized splitting pattern from $e_0$ to the distinguished tuple $e$] Define the partial type over the parameter set $\{e\}$ in the variable tuple $y$ by \begin{align*} \Gamma(y) := \{\theta_{m,n}(e,y): m,n\in\mathbb{N}\}. \end{align*} We claim that every finite subset of $\Gamma(y)$ is satisfiable in $S$. Indeed, let $\Gamma_0(y)\subset \Gamma(y)$ be finite. Then there are finitely many pairs $(m_1,n_1),\dots,(m_r,n_r)$ such that \begin{align*} \Gamma_0(y)=\{\theta_{m_\ell,n_\ell}(e,y):1\leq \ell\leq r\}. \end{align*} Since $S\models \theta_{m_\ell,n_\ell}(e_0,b_0)$ for each $\ell$, the formula \begin{align*} \exists y\,\bigwedge_{\ell=1}^r \theta_{m_\ell,n_\ell}(z,y) \end{align*} belongs to $\operatorname{tp}(e_0/\varnothing)$. Because both $e_0$ and $e$ realize $p(z)$, we have $\operatorname{tp}(e_0/\varnothing)=\operatorname{tp}(e/\varnothing)$. Hence \begin{align*} S\models \exists y\,\bigwedge_{\ell=1}^r \theta_{m_\ell,n_\ell}(e,y), \end{align*} so $\Gamma_0(y)$ is satisfiable in $S$. The parameter set $\{e\}$ has cardinality $|z|<\kappa$, and $\Gamma(y)$ has cardinality $\aleph_0<\kappa$. Since $S$ is $\kappa$-saturated, there is a tuple $b\in S^{|y|}$ such that \begin{align*} S\models \theta_{m,n}(e,b) \end{align*} for every $m,n\in\mathbb{N}$. [/step] [step:Contradict the finite-or-cofinite behavior in the saturated witness] Define \begin{align*} B := \{c\in \varphi(S,e): S\models \psi(c,b)\}. \end{align*} For every $m\in\mathbb{N}$, the formula $\theta_{m,1}(e,b)$ implies that $B$ has at least $m$ distinct elements. Therefore $B$ is infinite. For every $n\in\mathbb{N}$, the formula $\theta_{1,n}(e,b)$ implies that $\varphi(S,e)\setminus B$ has at least $n$ distinct elements. Therefore $\varphi(S,e)\setminus B$ is infinite. Thus $B$ is neither finite nor cofinite in $\varphi(S,e)$, contradicting the assumed finite-or-cofinite property of every $S$-definable subset of $\varphi(S,e)$. This contradiction shows that no such formula $\psi(x,y)$ and parameter tuple $a\in N^{|y|}$ exist. Since $M$, $e'$, and $N\succeq M$ were arbitrary, $\varphi(M,e')$ is strongly minimal in the stated sense. [/step]