How does the quantum teleportation protocol not violate the no-cloning theorem?

Consider the following protocol:

• Alice and Bob share the state
  begin{equation} |Phi^+rangle=frac{1}{sqrt{2}}(|0rangle|0rangle pm |1rangle|1rangle) end{equation}
• Alice has to teleport to Bob the state (which can be unknown even to her)
  begin{equation} |psirangle = c_0|0rangle + c_1|1rangle end{equation}
  So she appends this state to her part of the system in this way:
  begin{equation} |psirangleotimes|Phi^+rangle= frac{1}{sqrt{2}}(c_0|00rangle_A|0rangle_B+c_0|01rangle_A|1rangle_B+c_1|10rangle_A|0rangle_B+c_1|11rangle_A|1rangle_B)=end{equation}
  begin{equation} = frac{1}{2}(|Phi^+rangle_A(c_0|0rangle + c_1|1rangle)+|Phi^-rangle_A(c_0|0rangle – c_1|1rangle)+|Psi^+rangle_A(c_1|0rangle + c_0|1rangle)+|Psi^-rangle_A(c_1|0rangle – c_0|1rangle))end{equation}
  (where $|Psi^pmrangle$ and $|Phi^pmrangle$ are defined in the next stage of the protocol)
• Alice performs a Bell measurement, with projectors obtained from the Bell basis
  begin{equation} |Phi^pmrangle = frac{1}{sqrt{2}}(|0rangle|0rangle pm |1rangle|1rangle) end{equation}
  begin{equation} |Psi^pmrangle = frac{1}{sqrt{2}}(|0rangle|1rangle pm |1rangle|0rangle) end{equation}
• Alice sends two bits to Bob in order to communicate the the result of her measurement (e.g. 00 for $|Phi^+rangle$ and so on)
  -Now Bob applies a Pauli transformation on his part of the system depending on the result of the measurement of Alice and recovers the original state $|psirangle$.

Here comes the question: why this protocol DOES NOT violate the no-cloning theorem?