I have a problem with a proof I found in Velleman's "How to prove it". This is sort of interesting, because it is the very first time I cannot see the structure of a proof presented in the book. The following are the theorem and Velleman's proof.
Suppose $R$ is a partial order on a set $A$, and $B \subseteq A$. If $R$ is a total order and $b$ is a minimal element of $B$, then $b$ is the smallest element of $B$.
Suppose $R$ is a total order and $b$ is a minimal element of $B$. Let $x$ be an arbitrary element of $B$. If $x = b$, then since $R$ is reflexive, $bRx$. Now suppose $x \neq b$. Since $R$ is a total order, we know that either $xRb$ or $bRx$. But $xRb$ can’t be true, since by combining $xRb$ with our assumption that $x \neq b$ we could conclude that $b$ is not minimal, thereby contradicting our assumption that it is minimal. Thus, $bRx$ must be true. Since $x$ was arbitrary, we can conclude that $ \forall x \in B(bRx)$, so $b$ is the smallest element of $B$.
My problem is with the structure of the proof. In particular, I see a bit as a "rabbit out of the hat" the fact that we start by assuming that $x \neq b$.
In the pages of the book before the the actual proof, the author presents the scratch work behind the proof. Now, I can see the logic behind it (the all problem about the $x$ being or not being equal to $b$), but I do not see how this smoothly enters in the picture of the proof. Just consider that using "Proof Designer" (the software that can be found on Velleman's site and that goes along the book), I could actually prove the theorem without starting with the assumption that $x \neq b$. Indeed, the proof in words should go along the following lines:
Assume that $B \subseteq A$, that $R$ is a total order, and that $b$ is a minimal element of $B$. Let $x \in B$ be arbitrary. Thus, by the fact that $b$ is minimal, we have that if $(x,b) \in R$, then $x=b$. Proceed by cases.
1. $(x,b) \notin R$:
Since $x \in B$ and $b \in B$, we can conclude that $x \in A$ and $b \in A$. From the completeness of $R$ we have that either $(x,b) \in R$ or $(b,x) \in R$. Thus, by assumption that $(x,b) \notin R$, we have that $(b,x) \in R$.
Since $x \in B$ and $b \in B$, we can conclude that $x \in A$ and $b \in A$. From the reflexivity of $R$ we have that $(b,b) \in R$. But, since by assumption $x=b$, we can conclude that $(b,x) \in R$. QED
Is my proof sound?
Is there somebody who can give me a feedback on what is actually the structure of Velleman's proof?
I actually tried to replicated Velleman's structure of the proof with "Proof Designer", but I did not succeed.
Thanks a lot for any help.
PS: Conditional on my proof being sound, any feedback on the writing is more than wellcome.