Proof of Karp-Lipton theorem
Recall that $L \in \Sigma_i^p$ iff $\bar{L} \in \Pi_i^p$. Suppose now that $\Sigma_i^p \subseteq \Pi_i^p$, and let $L \in \Pi_i^p$. Then $\bar{L} \in \Sigma_i^p$ and so $\bar{L} \in \Pi_i^p$ by assumption, implying that $L \in \Sigma_i^p$. In other words, $\Pi_i^p \subseteq \Sigma_i^p$, and so $\Sigma_i^p = \Pi_i^p$.

Here's why $L \in \Sigma_i^p$ iff $\bar{L} \in \Pi_i^p$. For concreteness, we take $i = 3$. By definition, $L \in \Sigma_3^p$ if for some P-time predicate $T$,
$$
x \in L \Leftrightarrow \exists |y| &lt; |x|^{O(1)} \forall |z| &lt; |x|^{O(1)} \exists |w| &lt; |x|^{O(1)} T(x,y,z,w).
$$
Similarly $\bar{L} \in \Pi_3^p$ if for some P-time predicate $S$,
$$
x \in \bar{L} \Leftrightarrow \forall |y| &lt; |x|^{O(1)} \exists |z| &lt; |x|^{O(1)} \forall |w| &lt; |x|^{O(1)} S(x,y,z,w).
$$
However, these two statements are equivalent, as a simple invocation of de Morgan's laws shows, together with the fact that P is closed under complementation (take $S = \lnot T$).