Now we describe how we have calculated the cross sections of the above mentioned processes. From the quantized chiral Lagrangian that appears in eq.(\[eq:MENL\]) we obtain the corresponding Feynman rules (see Appendix). Then we compute, at tree level, the amplitudes for the different subprocesses contributing to the $Z^0Z^0$ or $W^{\pm}Z^0$ final state, using a REDUCE code, and then the corresponding cross-sections. We calculate them in the center of mass frame, with a total energy $\sqrt{\hat{s}}$. There, the four momenta of the interacting particles are:

$$\begin{aligned}
p_1 & = & (E_1, 0,0,p) \\
p_2 & = & (E_2, 0,0,p) \\
p_3 & = & (E_3,  p'{\rm sin}\theta , 0,  p'{\rm cos}\theta)   \\
p_4 & = & (E_4, -p'{\rm sin}\theta , 0, -p'{\rm cos}\theta)\end{aligned}$$

Here, $E_i$ are the particle energies, and $p$ and $p'$ are, respectively, the magnitudes of the three-momenta of the initial and the final states.

$$\begin{aligned}
p & = & \frac{1}{2\sqrt{\hat{s}}}{[{\hat{s}}^2
-2{\hat{s}}({m_1}^2+{m_2}^2)+({m_2}^2-{m_1}^2)^2]}^{1/2}  \\
p' & = & \frac{1}{2\sqrt{\hat{s}}}{[{\hat{s}}^2
-2{\hat{s}}({m_3}^2+{m_4}^2)+({m_4}^2-{m_3}^2)^2]}^{1/2}  \\
E_i & = & \sqrt{{m_i}^2+{p_i}^2} \;\;\;\;\;   i=1,..4  \\
{\hat{s}} & = & (p_1+p_2)^2    \\
{\hat{t}} & = & (p_1-p_3)^2    \\
{\hat{u}} & = & (p_1-p_4)^2\end{aligned}$$

where $\hat{s}$, $\hat{t}$ and $\hat{u}$ are the Mandelstam variables. The differential cross section is 