(This answer may look complex, but it really is simple in structure. I'm just going through things in baby steps.)
$ \newcommand{\odd}[1]{#1\text{ is odd}} \newcommand{\even}[1]{#1\text{ is even}} $Using slightly different notations (from Dijkstra/Scholten and Gries/Schneider), we are asked to prove that $ \left( \langle \cap i :: A_i \rangle \;\cap\; \langle \cup i : \odd i : A_i \rangle \right) \;\triangle\; \langle \cap i : \odd i : A_i \rangle $ and $ \left( \langle \cap i :: A_i \rangle \;\triangle\; \langle \cap i : \odd i : A_i \rangle \right) \;\cap\; \langle \cup i :: A_i \rangle $ are the same sets.
For situations like these, for me the simplest and most mechanical route is to expand the definitions and apply the rules of predicate logic to simplify the resulting expressions. Which $\;x\;$ are in the first set?
\begin{align} & x \in \left( \langle \cap i :: A_i \rangle \;\cap\; \langle \cup i : \odd i : A_i \rangle \right) \;\triangle\; \langle \cap i : \odd i : A_i \rangle \\ \equiv & \;\;\;\;\;\text{"definitions of $\;\cap,\triangle\;$"} \\ & \left( x \in \langle \cap i :: A_i \rangle \;\land\; x \in \langle \cup i : \odd i : A_i \rangle \right) \;\not\equiv\; x \in \langle \cap i : \odd i : A_i \rangle \\ \equiv & \;\;\;\;\;\text{"definitions of $\;\cap,\cup\;$ quantifications"} \\ & \left( \langle \forall i :: x \in A_i \rangle \;\land\; \langle \exists i : \odd i : x \in A_i \rangle \right) \;\not\equiv\; \langle \forall i : \odd i: x \in A_i \rangle \\ \equiv & \;\;\;\;\;\text{"logic: $\;\forall \Rightarrow \exists\;$ in left hand side of $\;\not\equiv\;$, assuming an odd $i$ exists"} \\ & \langle \forall i :: x \in A_i \rangle \;\not\equiv\; \langle \forall i : \odd i : x \in A_i \rangle \\ \equiv & \;\;\;\;\;\text{"logic: rewrite as detailed below"} \\ (*) \phantom\equiv & \langle \forall i : \odd i : x \in A_i \rangle \;\land\; \lnot \langle \forall i : \even i : x \in A_i \rangle \\ \end{align}
Here we used the following logical rewriting step:
\begin{align} & \langle \forall z :: P \rangle \;\not\equiv\; \langle \forall z : Q : P \rangle \\ \equiv & \;\;\;\;\;\text{"split range of left quantification -- to bring out the structural similarity"} \\ & \langle \forall z : Q : P \rangle \land \langle \forall z : \lnot Q : P \rangle \;\not\equiv\; \langle \forall z : Q : P \rangle \\ \equiv & \;\;\;\;\;\text{"bring $\;\lnot\;$ to outside"} \\ & \lnot \left( \langle \forall z : Q : P \rangle \land \langle \forall z : \lnot Q : P \rangle \;\equiv\; \langle \forall z : Q : P \rangle \right) \\ \equiv & \;\;\;\;\;\text{"$\;p \equiv p \land q\;$ is one way to write $\;p \Rightarrow q\;$"} \\ & \lnot \left( \langle \forall z : Q : P \rangle \;\Rightarrow\; \langle \forall z : \lnot Q : P \rangle \right) \\ \equiv & \;\;\;\;\;\text{"$\;\lnot p \lor q\;$ is another way to write $\;\Rightarrow\;$; DeMorgan"} \\ & \langle \forall z : Q : P \rangle \;\land\; \lnot \langle \forall z : \lnot Q : P \rangle \\ \end{align}
And similarly for the second set, we have for all $\;x\;$,
\begin{align} & x \in \left( \langle \cap i :: A_i \rangle \;\triangle\; \langle \cap i : \odd i : A_i \rangle \right) \;\cap\; \langle \cup i :: A_i \rangle \\ \equiv & \;\;\;\;\;\text{"definitions of $\;\triangle,\cap\;$"} \\ & \left( x \in \langle \cap i :: A_i \rangle \;\not\equiv\; x \in \langle \cap i : \odd i : A_i \rangle \right) \;\land\; x \in \langle \cup i :: A_i \rangle \\ \equiv & \;\;\;\;\;\text{"definitions of $\;\cap,\cup\;$ quantifications"} \\ & \left( \langle \forall i :: x \in A_i \rangle \;\not\equiv\; \langle \forall i : \odd i : x \in A_i \rangle \right) \;\land\; \langle \exists i :: x \in A_i \rangle \\ \equiv & \;\;\;\;\;\text{"logic: rewrite as detailed above"} \\ & \langle \forall i : \odd i : x \in A_i \rangle \;\land\; \lnot \langle \forall i : \even i : x \in A_i \rangle \;\land\; \langle \exists i :: x \in A_i \rangle \\ \equiv & \;\;\;\;\;\text{"logic: leftmost $\;\forall \Rightarrow \exists\;$, assuming an odd $i$ exists"} \\ (**) \phantom\equiv & \langle \forall i : \odd i : x \in A_i \rangle \;\land\; \lnot \langle \forall i : \even i : x \in A_i \rangle \\ \end{align}
Since $(*)$ and $(**)$ are identical, both sets have the same elements $\;x\;$, and therefore they are equal.
Note that we needed to assume that the index set of $\;i\;$ does not contain only even numbers. For example, if $\;i\;$ ranges over $\;\{4\}\;$, then the first set is 'the universe' and the second is $\;A_4\;$.