Is this proof valid? I def feel like there is a much shorter solution that I missed.
Theorem. Let $E$ be the set of all equiv relations on $\mathbb{Z}^+$. Then $|E|=|P(\mathbb{Z}^+)|$.
Proof:
Given that $^{\mathbb{Z}^+}\mathbb{Z}^+$, the set of all functions $\mathbb{Z}^+\to\mathbb{Z}^+$, is equinumerus with $P(\mathbb{Z}^+)$, we construct injections $E\to^{\mathbb{Z}^+}\mathbb{Z}^+$ and $P(\mathbb{Z}^+)\to E$; which then implies a bijection $E\to P(\mathbb{Z}^+)$.
Injection $\displaystyle E\rightarrow ^{Z^{+}} Z^{+}$:
Let $h:E\rightarrow ^{Z^{+}} Z^{+}$ be defined such that $h(R)$ is the function $g:Z^{+}\rightarrow Z^{+}$ given by $g(x)=\min [x]_{R}$, where $\min [x]_{R}$ denotes the smallest element in the equivalence class $[x]_{R}$. To prove that $h$ is injective, assume for the sake of contradiction that $h(R_{1} )=h(R_{2} )$ for two distinct equivalence relations $R_{1}$ and $R_{2}$. This implies that $\min [x]_{R_{1}} =\min [x]_{R_{2}}$ for all $x$ in $Z^{+}$. Since $R_{1} \neq R_{2}$, there must exist elements $x,y$ such that $y\in [x]_{R_{1}}$ but $y\notin [x]_{R_{2}}$. Consequently, $\min [y]_{R_{2}} =\min [y]_{R_{1}} =\min [x]_{R_{1}} =\min [x]_{R_{2}}$. However, this is a contradiction because $[y]_{R_{2}} \neq [x]_{R_{2}}$, and therefore $\min [y]_{R_{2}}$ cannot equal $\min [x]_{R_{2}}$. This contradiction establishes that $h$ is injective.
Injection $\displaystyle P\left( Z^{+}\right)\rightarrow E$:
Let $\displaystyle h:P\left( Z^{+}\right)\rightarrow E$ be defined by sending $\displaystyle X\subseteq Z^{+}$ to the equiv relation $\displaystyle R$ defined by: $\displaystyle x\ R\ y$ iff $\displaystyle x,y\in X$ AND have same prime factors OR $\displaystyle x,y\notin X$; for simplicity, suppose $\displaystyle 1$ is not in $\displaystyle Z^{+}$, so we're only considering composites and primes. To see that this map $\displaystyle h$ is injective, let $\displaystyle X_{1} ,X_{2} \subseteq Z^{+}$ with $\displaystyle X_{1} \neq X_{2}$. Then $\displaystyle x\in X_{1}$, $\displaystyle x\notin X_{2}$ for some $\displaystyle x$. Let $\displaystyle h( X_{1}) =R_{1}$ and $\displaystyle h( X_{2}) =R_{2}$. Now, if there exists $\displaystyle y\in Z^{+}$ such that $\displaystyle y\notin X_{1}$ and $\displaystyle y\notin X_{2}$, then $\displaystyle x,y\notin X_{2}$, so $\displaystyle x\ R_{2} \ y$. We also have $\displaystyle ( x,y) \notin R_{1}$, so $\displaystyle R_{2} \neq R_{1}$.
Otherwise, $\displaystyle X_{1} \cap X_{2} =Z^{+}$, so $\displaystyle X_{2} =Z^{+} \setminus X_{1}$. Without loss of generality let $\displaystyle x$ and $\displaystyle y$ be two primes in $\displaystyle X_{1}$. Then $\displaystyle \neg x\ R_{1} \ y$ but also $\displaystyle x\ R_{2} \ y$, since $\displaystyle x,y\notin X_{2}$. In both cases we get $\displaystyle h( X_{1}) \neq h( X_{2})$, so $\displaystyle h$ is injective.