We know that the satisfiability problem for a formula in the form of $\exists x_0 \forall x_1 \exists x_2 \ldots Q_i x_i . \phi(x_0, \ldots, x_i)$ is complete for $\Sigma_{i}^p$, where $Q_i$ is a quantifier $\exists$ iff $i \equiv 0 \mod 2$, $\forall$ otherwise, hence there are $i$ quantifier alternations. Similarly, a $\Pi_i^p$-complete satisfiability problem can be defined. Additionally, $\Sigma_{i}^p \subseteq \Pi_{i+1}^p$ and $\Pi_{i}^p \subseteq \Sigma_{i+1}^p$.
We also know that $\exists x . \phi(x) \equiv \phi(\top) \vee \phi(\bot)$, and $\forall x. \phi(x) \equiv \phi(\top) \wedge \phi(\bot)$. Moreover, the elimination of the first quantifier can be done in logarithmic space.
To my understanding, this is a reduction of a $\Sigma_{i+1}^p$-complete problem to a problem in $\Pi_{i}^p$, as it removes 1 quantifier alternation and is computable in logarithmic space. But that would conclude that $\Sigma_{i+1}^p \subseteq \Pi_{i}^p$ and $\Pi_{i+1}^p \subseteq \Sigma_{i}^p$, implying the collapse of $PH$. So:
Where am I wrong?