Inequalities proofs in Spivak's Calculus From Chapter 1, page 9 of Spivak's Calculus, 3rd edition:

In fact, it is convenient to consider the collection of all positive numbers, denoted by $P$, as the basic concept, and state all properties in terms of $P$:
(P10) (Trichotomy law) For every number $a$, one and only one of the following holds:
(i) $a=0$, (ii) $a$ is in the collection $P$, (iii) $-a$ is in the collection $P$.
(P11) (Closure under addition) If $a$ and $b$ are in $P$, then $a+b$ is in $P$.
(P12) (Closure under multiplication) If $a$ and $b$ are in $P$, then $a\cdot b$ is in $P$.

This is question 8 on page 15:

Although all the basic properties of inequalities were stated in terms of the collection $P$ of all positive numbers, and < was defined in terms of $P$, this procedure can be reversed. Suppose that P10-P12 are replaced by
(P'10) For any numbers $a$ and $b$ one, and only one, of the following holds: (i) $a=b$, (ii) $a<b$, (iii) $b<a$.
(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 $ac<bc$.
Show that P10-P12 can then be deduced as theorems.

And here's the solution as in the manual:

Two applications of P'12 show that if $a<b$ and $c<d$, then $a+c<b+c<b+d$, so $a+c<b+d$ by P'11. In particular, if $0<b$ and $0<d$, then $0<b+d$, which proves P11. It follows, in addition, that if $a<0$, then $-a>0$; for if $-a<0$ were true, then $0=a+(-a)<0$, contradicting P'10. Consequently, any number $a$ satisfies precisely one of the conditions $a=0$, $a>0$, $a<0$, the last being equivalent to $-a>0$. This proves P10. Finally, P'13 shows that if $0<a$ and $0<c$, then $0<ac$, which proves P12.

My question is, why doesn't P10 immediately follow from P'10, by substituting $0$ for $b$ ?
 A: Because then what we would have would be:

For any number $a$, one, and only one, of the following holds: (i) $a=0$, (ii) $a<0$, (iii) $0<a$.

It is not obvious that you can jump from this to:

For any number $a$, one, and only one, of the following holds: (i) $a=0$, (ii) $0<-a$, (iii) $0<a$.

You will need some connection between $<$ and the arithmetic operations in $\Bbb R$.
A: I had the same thought about this problem. It's easy to be confused about which properties still hold.
The key point is that here, we aren't allowed to use the equivalence of $a<0$ and $-a>0$. This equivalence is actually what Spivak's proving.
See if you can follow how his proof shows this.
Once he has this, he just sticks it into 'P10 to complete the proof.
As an aside, his explanation seems to skip a possibility: he rules out $-a <0$ but not $-a=0$.
This can be ruled out by assuming it's true and then adding $-a$ to both sides of $a<0$, per 'P12. (It's the same argument he uses to rule out $-a <0$).
Finally, note that Spivak's argument for why $a< 0$ implies $-a > 0$ is perhaps needlessly circuitous.
An alternate, more direct approach:
Begin as Spivak does with $a < 0$.
Add $-a$ to both sides per 'P12.
