Without loss of generality, suppose $f(x_1) = x_2$, $f(x_2) = x_3$, $f(x_3) = x_1$ with $x_1 < x_2 < x_3$. We construct the horseshoe for $f^2$ in steps. **Step 1: Existence of a fixed point.** On the interval $[x_2, x_3]$: $f(x_2) - x_2 = x_3 - x_2 > 0$ and $f(x_3) - x_3 = x_1 - x_3 < 0$. By the intermediate value theorem, there exists $z \in (x_2, x_3)$ with $f(z) = z$. **Step 2: Existence of a preimage $y$ of $z$ in $(x_1, x_2)$.** On $[x_1, x_2]$: since $z \in (x_2, x_3)$ we have $f(x_1) = x_2 < z$, giving $f(x_1) - z < 0$, while $f(x_2) = x_3 > z$, giving $f(x_2) - z > 0$. By the intermediate value theorem there exists $y \in (x_1, x_2)$ with $f(y) = z$. **Step 3: Key identity for $f^2$.** Since $f(y) = z$ and $f(z) = z$, we have $f^2(y) = f(z) = z$ and $f^2(z) = f(z) = z$. **Step 4: Existence of smallest $r \in [y, x_2)$ with $f^2(r) = y$.** We have $f^2(y) = z > y$ and $f^2(x_2) = f(x_3) = x_1 < y$. By the intermediate value theorem there exists $r \in (y, x_2)$ with $f^2(r) = y$; take $r$ to be the smallest such value. **Step 5: Existence of largest $s \in (x_2, z)$ with $f^2(s) = y$.** We have $f^2(z) = z > y$ and we need a point near $x_2$ from the right where $f^2$ dips below $y$. At $x_2$: $f^2(x_2) = x_1 < y$. Applying the intermediate value theorem on $(x_2, z)$ gives $s \in (x_2, z)$ with $f^2(s) = y$; take $s$ to be the largest. **Conclusion.** We have verified: \begin{align*} f^2(y) = z, \quad f^2(r) = y, \quad f^2(s) = y, \quad f^2(z) = z. \end{align*} Set $K_1 = (y, r)$, $K_2 = (s, z)$, $J = (y, z)$. Then $K_1, K_2 \subset J$ are disjoint open subintervals and $f^2(K_1) = J$, $f^2(K_2) = J$, because $f^2$ maps each of $[y,r]$ and $[s,z]$ onto $[y,z]$ by continuity and the boundary values computed above. This is a horseshoe for $f^2$, so $f$ is Glendinning-chaotic.