What is wrong with this putative proof? So I've spent about an hour trying to figure out what is wrong with this proof. Could somebody clearly explain it to me? I don't need a counterexample. For some reason I was able to figure that out.
Thanks.
Theorem. $\;$ Suppose $R$ is a total order on $A$ and $B\subseteq A$. Then every element of $B$ is either the smallest element of $B$ or the largest element of $B$.
Proof. $\;$ Suppose $b\in B$. Let $x$ be an arbitrary element of $B$. Since $R$ is a total order, either $bRx$ or $xRb$. 


*

*Case 1. $bRx$. Since $x$ was arbitrary, we can conclude that $\forall x\in B(bRx)$, so $b$ is the smallest element of $R$.

*Case 2. $xRb$. Since $x$ was arbitrary, we can conclude that $\forall x\in B(xRb)$. so $b$ is the largest element of $R$.


Thus, $b$ is either smallest element of $B$ or the largest element of $B$. Since $b$ was arbitrary, every element of $B$ is either its smallest element or its largest element.

The proof shows that for an arbitrary $x$, either $bRx$ or $xRb$. Therefore $\forall x(xRb\lor bRx)$. So far so good. 
But $\forall x(P(x)\lor Q(x))$ is not equivalent to $\forall xP(x)\lor\forall x Q(x)$.
