## Appendix A Proofs of Theorems and Corollaries
We must produce \(x\in\Lambda\) with \(S(x)=s\). Note that if \(x\in\cap_{0\leq n\leq\infty}I_{s_{0}s_{1}\ldots s_{n}}\), then \(x\in I_{s_{0}},f(x)\in I_{s_{1}}\), etc. Hence, there exists \(x\in\Lambda\) such that \(S(x)=(s_{0}s_{1}s_{1}\ldots)\). This proves that \(S\) is onto. Observe that \(x\in\cap_{0\leq n\leq\infty}I_{s_{0}s_{1}\ldots s_{n}}\) consists of a unique point because \(S\) is one to one. In particular, we have that diameter of \(I_{s_{0}s_{1}\ldots s_{n}}\) tending to \(0\) as \(n\to\infty\). Furthermore, to prove the continuity of \(S\), we choose \(x\in\Lambda\) and suppose that \(S(x)=(s_{0}s_{1}s_{2}\ldots)\). Let \(\epsilon>0\). Pick \(n\) so that \(1/2^{n}<\epsilon\). Consider the closed subintervals \(I_{s_{0}t_{1}\ldots t_{n}}\) defined above for all possible combinations \(t_{0}t_{1}\ldots t_{n}\). These subintervals are all disjoint, and \(\Lambda\) is contained in their union. These are \(2^{n+1}\) such intervals, and \(I_{s_{0}s_{1}\ldots s_{n}}\) is one of them. Hence, we may choose \(\delta\) such that \(|x-y|<\delta\) and \(y\in\Lambda\) implies that \(y\in I_{s_{0}s_{1}\ldots s_{n}}\). Therefore, \(S(y)\) agrees with \(S(x)\) in the first \(n+1\) terms. Thus, by Proposition A2, we have \(d(S(x),S(y))\leq\frac{1}{2^{n}}<\epsilon\). This proves the continuity of \(S\). Since \(\Lambda\) is compact (bounded closed sets) and \(S\) is continuous and one to one, \(S^{-1}\) is also continuous. Thus, \(S\) is a homeomorphism. Finally, to prove that \(S\circ f=\sigma\circ S\), i.e., \(S\) provides an equivalence between the dynamics of \(f\) on \(\Lambda\) and \(\sigma\) on \(\Sigma_{2}\). Let \(x_{0}\in\Lambda\) and suppose that \(S(x_{0})=(s_{0}s_{1}s_{2}\ldots)\). Then, we have \(x_{0}\in I_{z_{0}},f(x_{0})\in I_{z_{1}},f^{2}(x_{0})\in I_{z_{2}}\) and so forth. However, the fact that \(f(x_{0})\in I_{z_{1}},f^{2}(x_{0})\in I_{z_{2}}\), etc., says that \(S(f(x_{0}))=(s_{1}s_{2}s_{3}\ldots)\), so \(S(f(x_{0}))=\sigma(S(x_{0}))\), which is what we wanted to prove. The proof is now complete. Proof of Theorem 4.: Let \(\mathcal{R}\) be the set of real numbers. Under the conditions that Theorem 4 holds, it is easy to check the following properties: (1) the break points \(-\frac{(k_{1}-n_{1})k_{1}}{n_{1}-1}(<0)\) and \(-\frac{(k_{2}-n_{1})k_{1}}{n_{1}-1}(<0)\) are given by the solutions of \(\{x\in\mathcal{R}|H_{1}(x)=k_{1}\}\) and \(\{x\in\mathcal{R}|H_{1}(x)=k_{2}\}\), respectively. (2) The break points of \(H(x_{n})\): \(k_{1}\) and \(k_{2}\) are also the break points of \(H^{2}(x_{n})\). (3)The break points \(k_{2}-k_{2}(k_{2}-k_{1})\) and \(k_{2}(1-k_{1})+k_{1}^{2}\) are given by the solutions of \(\{x\in\mathcal{R}|H_{2}(x)=k_{2}\}\) and \(\{x\in\mathcal{R}|H_{2}(x)=k_{1}\}\), respectively. (4) \(0<k_{1}<k_{2}-k_{2}(k_{2}-k_{1})<k_{2}(1-k_{1})+k_{1}^{2}<k_{2}\) holds because \(k_{2}-(k_{2}(1-k_{1})+k_{1}^{2})=k_{1}(k_{2}-k_{1})>0\), \(k_{2}(1-k_{1})+k_{1}^{2}-(k_{2}-k_{2}(k_{2}-k_{1}))=(k_{2}-k_{1})^{2}>0\) and \(k_{2}-k_{2}(k_{2}-k_{1})-k_{1}=(1-k_{2})(k_{2}-k_{1})>0\) hold. (5) The break points \(k_{2}+\frac{k_{1}(1-k_{2})}{n_{2}}\) and \(k_{2}+\frac{k_{2}(1-k_{2})}{n_{2}}\) are given by the solutions of \(\{x\in\mathcal{R}|H_{3}(x)=k_{1}\}\) and \(\{x\in\mathcal{R}|H_{3}(x)=k_{2}\}\), respectively. (6) \(k_{2}<k_{2}+\frac{k_{1}(1-k_{2})}{n_{2}}<1<k_{2}+\frac{k_{2}(1-k_{2})}{n_{2}}\) holds because \(k_{2}+\frac{k_{2}(1-k_{2})}{n_{2}}-1=(1-k_{2})(\frac{k_{2}}{n_{2}}-1)>0\) and \(1-(k_{2}+\frac{k_{1}(1-k_{2})}{n_{2}})=(1-k_{2})(1-\frac{k_{1}}{n_{2}})>0\) hold. Furthermore, \(|DH_{1}H_{1}|<1\) because \(0<(1-n_{1}),k_{1}<1\) and \(1-n_{1}<k_{1}\). \(|DH_{2}H_{2}|>1\) because \(0<k_{1},k_{2}<1\), \(k_{1}<k_{2}\), and \(0<k_{2}-k_{1}<1\). \(|DH_{3}H_{3}|<1\) because \(0<(1-k_{2}),n_{2}<1\) and \(k_{2}<1-n_{2}\). \(|DH_{3}H_{1}|=|DH_{1}H_{3}|<1\) because \(1-n_{1}<k_{1}\) and \(k_{2}<1-n_{2}\). \(|DH_{3}H_{2}|=|DH_{2}H_{3}|>1\) because \(0<k_{2}-k_{1}<1\), and \(n_{2}>(k_{2}-k_{1})(1-k_{2})\). \(|DH_{1}H_{2}|=|DH_{2}H_{1}|>1\) because \(0<k_{2}-k_{1}<1\) and \(1-k_{1}(k_{2}-k_{1})>n_{1}\).