✳204.7 $\vdash: P \in Ser .\supset. P_1 \in 1 \rightarrow 1$
Which says if $P$ is a series, then $P_1$ is one-one.
✳201.63 $\vdash: P \in trans \cap Rl‘J .\supset. P_1 = P \overset{.}{-}P^2$
Which says if $P$ is transitive and asymetrical, then $P_1=P\overset{.}{-}P^2$
Where
$P_1$ is "immediately precedes" as defined in ✳121.02 :
✳121.02 $P_v = \hat{x} \hat{y}\{ N_0c‘P(x|-|y)=v+_c1 \}$ Df where $v=1$
and $P^2 = P|P$ as defined in ✳34.02
Let R be "less than confined to the real interval [0,1]," then R is transitive, asymmetrical and connected. Therefore, by definition, $R$ is a series. But there is no term in $R$'s field that immediately precedes any other terms. In other words, $R_1$ does not seem to exist. If I'm not mistaken, the premises of 204.7 should include $\overset{.}{\exists}!P_1$ as a hypothesis.

If I understood well the definition, it is trivial that in $[0,1]$ there is an element which precedes any other element : it is $0$.
But to say that for any element different from $0$ there is an immediate predecessor, we have to use something like the Well-ordering Theorem proved by Zermelo in 1904 (thus known to W&R at time of writing PM) using the Axiom of Choice (the Multiplicative Axiom in PM).
This theorem says that any set can be well-ordered, and so also $[0,1]$.
But the proof is non-constructive: thus it gives us no clue about the way of finding the immediate predecessor of (say) $1$.