This is an exercise from Spivak's "Calculus". I am confused about how to prove that one, and only one, of three properties hold for an arbitrary number.
- (P'10) For any numbers $a$ and $b$ one, and only one, of the following holds:
- $a=b$
- $a < b$
- $b < a$
- (P'11) For any numbers $a$, $b$, and $c$, if $a < b$ and $b < c$, then $a < c$.
- (P'12) For any numbers $a$, $b$, and $c$, if $a < b$, then $a+c < b + c$.
- (P'13) For any numbers $a$, $b$, and $c$, if $a < b$ and $0 < c$, then $ac < bc$.
Given the properties above, prove that for every number $a$, one, and only one, of the following holds:
- $a=0$
- $a$ is in the collection $P$
- $-a$ is in the collection $P$
Labeling the three properties as $x$, $y$, and $z$ respectively, would the correct logical form for this proof be $(x \rightarrow \neg(y \vee z)) \wedge (y \rightarrow \neg(x \vee z)) \wedge (z \rightarrow \neg(x \vee y))$? Simplifying this gives $(x \rightarrow (\neg y \wedge \neg z)) \wedge (y \rightarrow \neg z)$.
Proof: Let $a$ be an arbitrary number. Suppose $a = 0$. By P'10, it follows that $a \geq 0$ and $0 \geq a$. Suppose that $a$ is in $P$. Then $a > 0$, which contradicts the fact that $0 \geq a$. So $a$ is not in $P$. Similarly, we can show that $-a$ is not in $P$.
Now suppose that $a$ is in $P$, so $a > 0$. Suppose that $-a$ is in $P$. Then $-a > 0$. By P'12, we have $0+(-a) < a + (-a) = 0$, which is a contradiction. Thus, $-a$ is not in P. $\square$