[proofplan] The proof is a direct translation through the assumed equivalence between the variational and DLR descriptions. We take two distinct equilibrium states for $\beta$, use the hypothesis to show that both satisfy the DLR equations for $\Phi$, and observe that distinct measures remain distinct under this identification. The definition of phase transition is then exactly the non-uniqueness of DLR Gibbs states. [/proofplan] [step:Introduce the two measure classes being identified] Let $\mathcal{M}_\sigma(X)$ denote the set of shift-invariant Borel probability measures on $X$. Define $\mathcal{E}_\beta \subset \mathcal{M}_\sigma(X)$ to be the set of equilibrium states for the continuous potential $\beta$, and define $\mathcal{G}_\Phi^\sigma \subset \mathcal{M}_\sigma(X)$ to be the set of shift-invariant DLR Gibbs states for the finite-range interaction $\Phi$. The hypothesis says precisely that, for every $\mu \in \mathcal{M}_\sigma(X)$, one has $\mu \in \mathcal{E}_\beta$ if and only if $\mu \in \mathcal{G}_\Phi^\sigma$. Hence the two subsets of $\mathcal{M}_\sigma(X)$ are equal: \begin{align*} \mathcal{E}_\beta = \mathcal{G}_\Phi^\sigma. \end{align*} [/step] [step:Transfer distinct equilibrium states to distinct DLR Gibbs states] By assumption, there exist measures $\mu_1, \mu_2 \in \mathcal{M}_\sigma(X)$ such that $\mu_1 \ne \mu_2$ and $\mu_1, \mu_2 \in \mathcal{E}_\beta$. Since $\mathcal{E}_\beta = \mathcal{G}_\Phi^\sigma$, it follows that $\mu_1, \mu_2 \in \mathcal{G}_\Phi^\sigma$. Therefore $\Phi$ has at least two distinct shift-invariant DLR Gibbs states. [guided] We begin with the exact meaning of the hypothesis that $\beta$ has at least two shift-invariant equilibrium states. It gives two shift-invariant Borel probability measures $\mu_1$ and $\mu_2$ on $X$ such that $\mu_1 \ne \mu_2$, with both measures attaining the variational equilibrium condition for $\beta$. In the notation introduced above, this says \begin{align*} \mu_1, \mu_2 \in \mathcal{E}_\beta \end{align*} and also $\mu_1 \ne \mu_2$ as Borel probability measures on $X$. The assumed correspondence is not a quotient or a map that might identify different measures. It is an equivalence statement about the same measure $\mu$: a measure belongs to $\mathcal{E}_\beta$ exactly when that same measure belongs to $\mathcal{G}_\Phi^\sigma$. Therefore membership transfers directly. Since $\mu_1 \in \mathcal{E}_\beta$, we get $\mu_1 \in \mathcal{G}_\Phi^\sigma$; since $\mu_2 \in \mathcal{E}_\beta$, we get $\mu_2 \in \mathcal{G}_\Phi^\sigma$. The distinction between the two measures is also preserved, because no transformation of measures has been applied. The two DLR Gibbs states obtained are exactly $\mu_1$ and $\mu_2$, and the original inequality $\mu_1 \ne \mu_2$ remains true. Thus $\mathcal{G}_\Phi^\sigma$ contains at least two distinct elements. [/guided] [/step] [step:Conclude non-uniqueness of Gibbs states] By definition, the finite-range interaction $\Phi$ has a phase transition when its DLR Gibbs states are not unique. Since $\mathcal{G}_\Phi^\sigma$ contains the two distinct measures $\mu_1$ and $\mu_2$, uniqueness fails even within the shift-invariant DLR Gibbs states. Consequently $\Phi$ has a phase transition. [/step]