Not a duplicate of
Proof verification: $P(A\cup B)=P(A)\cup P(B)\rightarrow A\subseteq B\vee A\supseteq B$
How do you prove $P(A) \cup P(B) = P(A \cup B) \Rightarrow (A \subseteq B) \lor (B \subseteq A)$
This is exercise $3.5.8$ from the book How to Prove it by Velleman $($$2^{nd}$ edition$)$:
Prove that for any sets $A$ and $B$, if $\mathscr P(A)\cup\mathscr P(B)=\mathscr P(A\cup B)$ then either $A\subseteq B$ or $B\subseteq A$.
Here is my proof:
Let $A$ and $B$ be arbitrary sets. Suppose $\mathscr P(A)\cup\mathscr P(B)=\mathscr P(A\cup B)$. Now we consider two different cases.
Case $1.$ Suppose $A\subseteq B$. Ergo $A\subseteq B$ or $B\subseteq A$.
Case $2.$ Suppose $A\nsubseteq B$. So we can choose some $x_0$ such that $x_0\in A$ and $x_0\notin B$. Let $y$ be an arbitrary element of $B$. Since $A\cup B\in\mathscr P(A\cup B)$, then $A\cup B\in\mathscr P(A)\cup\mathscr P(B)$. So either $A\cup B\subseteq A$ or $A\cup B\subseteq B$. Again we consider two different cases.
Case $2.1.$ Suppose $A\cup B\subseteq A$. Since $y\in B$, $y\in A\cup B$. Ergo $y\in A$.
Case $2.2.$ Suppose $A\cup B\subseteq B$. Since $x_0\in A$, $x_0\in A\cup B$. Ergo $x_0\in B$ which is a contradiction.
From $y\in A$ or a contradiction we obtain $y\in A$. Thus if $y\in B$ then $y\in A$. Since $y$ is arbitrary, $\forall y(y\in B\rightarrow y\in A)$ and so $B\subseteq A$. Ergo $A\subseteq B$ or $B\subseteq A$.
Since case $1$ and case $2$ are exhaustive, $A\subseteq B$ or $B\subseteq A$. Therefore if $\mathscr P(A)\cup\mathscr P(B)=\mathscr P(A\cup B)$ then either $A\subseteq B$ or $B\subseteq A$. Since $A$ and $B$ are arbitrary, $\forall A\forall B\Bigr(\mathscr P(A)\cup\mathscr P(B)=\mathscr P(A\cup B)\rightarrow(A\subseteq B\lor B\subseteq A)\Bigr)$. $Q.E.D.$
Is my proof valid$?$
Thanks for your attention.