Proposition 1. If \(a>b\) and \(b>c\), then \(a>c\).

Proof. Our assumptions that \(a>b\) and \(b>c\) imply that \(a-b\in\mathbb{P}\) and \(b-c\in\mathbb{P}\). Since \(\mathbb{P}\) is closed under addition, we have

\[a-c = (a-b)+(b-c)\in\mathbb{P}.\]

By definition, this means \(a>c\), as desired. \(\Box\)

We use the notation \(a<b<c\) to mean \(a<b\) and \(b<c\). Similarly for sybols such as \(a\leq b<c\). We don't use things like \(a<b>c\), where the inequalities don't go in the same direction.

Proof. By assumption \(a-b>0\), and hence

\[0<a-b = (a+c) - (b+c).\]

Hence, \((a+c)-(b+c)>0\), which implies \(a+c>b+c\). \(\Box\)

Proof. Since \(a-b\) and \(c\) are in \(\mathbb{P}\), and \(\mathbb{P}\) is closed under multiplication, we see that

\[ca-cb = c(a-b)\in\mathbb{P}.\]

We conclude that \(ca>cb\). \(\Box\)

Proof. By assumption \(b-a>0\) and \(c-b\geq 0\). If \(c-b=0\) then \(c=b\) and hence \(c-a=b-a>0\). Therefore, \(a<c\). On the other hand, if \(c-b>0\), then by Proposition 1 we can conclude that \(c>a\). \(\Box\)