I think you're making the first proof harder to follow than it needs to be. You already know that $P, -P, \{0 \}$ divide Gaul (well, your field $F$) into three disjoint parts. Therefore, you know that $b-a$ falls into exactly one of those parts. We know that $b-a=0$ occurs exactly when $b=a$, and in that case $0=b-a=a-b \notin P$, so $b=a \Rightarrow \lnot(a \lt b \lor b \lt a)$.

Otherwise, either $b-a \in P$ or $b-a \in -P$. This latter alternative occurs exactly when $-(b-a)=a-b \in P$. That proves trichotomy because now we know that for $b \neq a, b-a$ and $a-b$ can't both occur in $P$, but one of them must occur in $P$.

The remainder of your proof looks fine.

I think you're making the first proof harder to follow than it needs to be. You already know that $P, -P, \{0 \}$ divide Gaul (well, your field $F$) into three disjoint parts. Therefore, you know that $b-a$ falls into exactly one of those parts. We know that $b-a=0$ occurs exactly when $b=a$, and in that case $0=b-a=a-b \notin P$, so $b=a \iff \lnot(a \lt b \lor b \lt a)$.

Otherwise, either $b-a \in P$ or $b-a \in -P$. This latter alternative occurs exactly when $-(b-a)=a-b \in P$. That proves trichotomy because now we know that for $b \neq a, b-a$ and $a-b$ can't both occur in $P$, but one of them must occur in $P$.

The remainder of your proof looks fine.