Related to: Deriving the confidence interval $P(-\Phi^{-1}_\alpha < X < \Phi^{-1}_\alpha) = 1-2\alpha$
I'm not sure why, but I'm having some trouble with the definitions here.
$$P\sim\mathcal N(0,1)$$ \begin{align*} &\alpha &=&\quad P(X>z_\alpha)&(1)\\ &P(X>b) &=&\quad 1-P(X<b)&(2)\\ &P(X<c) &=&\quad \Phi(c)&(3)\\ &z_d &=&\quad \Phi^{-1}(d)&(4)\\ &f(f^{-1}) &=&\quad \mathrm{id}&(5) \end{align*} \begin{align*} &(1) \text{ and }(4) \implies &&\alpha = P(X>\Phi^{-1}(\alpha))&&(6)\\ &(6) \text{ and }(2) \implies &&\alpha = 1-P(X<\Phi^{-1}(\alpha))&&(7)\\ &(7) \text{ and }(3) \implies &&\alpha = 1-\Phi(\Phi^{-1}(\alpha))&&(8)\\ &(8) \text{ and }(5) \implies &&\alpha = 1-\alpha&&(9) \end{align*}
(9) seems to imply that $\alpha$ is a fixed constant, while (1) claims it's a variable, so why is this result wrong?
According to (1), (2) and (3) we have:$$\alpha=P(X>z_{\alpha})=1-P(X<z_{\alpha})=1-\Phi(z_{\alpha})$$ so that: $$1-\alpha=\Phi(z_{\alpha})$$ and consequently: $$\Phi^{-1}(1-\alpha)=z_{\alpha}$$ This contradicts (4).