Defining the polynomial hierarchy using $\Pi_i^p$ instead of $\Sigma_i^p$
It is a direct consequence of the definition of $\textsf{PH}$.

Remember:

$\qquad \Sigma_i^P = \textsf{NP}^{\Sigma_{i-1}^P}$

$\qquad \Pi_i^P = \textsf{coNP}^{\Sigma_{i-1}^P} $

What we want to prove is:

$\qquad \bigcup_{i&gt;0}\Sigma_i^P = \bigcup_{i&gt;0}\Pi_i^P $

We have:

$\qquad \Sigma^P_i \subseteq \textsf{coNP}^{\Sigma^P_i} = \Pi_{i+1}^P$

$\qquad \Pi^P_i \subseteq \textsf{NP}^{\Pi^P_i} = \textsf{NP}^{\Sigma_i^P} = \Sigma^P_{i+1}$

Taking the union yields:

$\qquad \bigcup_{i&gt;0} \Sigma^P_{i} \subseteq \bigcup_{i&gt;0} \Pi^P_{i} \subseteq \bigcup_{i&gt;0} \Sigma^P_{i}$

which is exactly our goal.