Note that this implies that $z=x$ and $z'=x'$, so that $z'=x'=e_jx = e_jz$. We also have $\varepsilon_i(x)=\varepsilon_i(\psi_i(x))$, which equals the maximum of $\varepsilon_i(y)$ and $t+\varepsilon_i(y)-\varphi_i(y)$. In particular $\varepsilon_i (y) = 0$ and by similar arguments $\varepsilon_i(y')=0$. Since $\varepsilon_i^\star(x)=t=t'=\varepsilon_i^\star(x')$ and $\varepsilon_i (x') = \varepsilon_i (x) = 0$, (14.8) implies (14.7) provided we can show $\varphi_i(y') - \varphi_i(y) = \varphi_i^\star(z') - \varphi_i^\star(z)$. But $y'=e_j y$, $z' = e_jz$ and in addition $\varepsilon_i(y)=\varepsilon_i(y')$ and $\varepsilon_i^\star(z) = \varepsilon_i^\star(x)=\varepsilon_i^\star(x') = \varepsilon_i^\star(z')$, so both $\varphi_i(y') - \varphi_i(y)$ and $\varphi_i^\star(z') - \varphi_i^\star(z)$ equal $\langle\alpha_j,\alpha_i^\vee\rangle$.