For any ordinal $\alpha$, we call any function $\alpha\rightarrow\{-,+\}$ a surreal number.
Denote by $D(x)$ the domain or Day of $x$, and by $\delta(x,y)$ the smallest ordinal for which $x$ and $y$ differ. We can define an order on the surreal numbers by saying $x<y$ iff (A) $\delta(x,y)=D(x)<D(y)$ and $y(\delta(x,y))=+$ OR (B) $\delta(x,y)=D(y)<D(x)$ and $x(\delta(x,y))=-$ OR (C) $\delta(x,y)<D(x),D(y)$ and $x(\delta(x,y))=-$ and $y(\delta(x,y))=+$.
I'm trying to prove $x<y \wedge y < z \Rightarrow x < z$ and going through this case by case, but I stumbled. I'm currently assuming that $x<y$ satisfies (A) and $y<z$ satisfies (B). If I further assume that $x$ is simpler than $z$ (i.e. $D(x)<D(z)$ and for $\beta\in D(x)$ holds $x(\beta)=z(\beta)$, I can prove it. For example take $x=+,y=++--,z=++-$. In the same way, if I assume that $z$ is simpler than $x$, I can prove it. For example take $x=+--+,y=+--++,z=+-$.
But I cannot solve the case that neither is simpler than the other and I cannot think of an example for that either. Probably it isn't possible, but I can't figure it out.
EDIT: I figured that part out. If $x$ and $z$ are simpler than $y$ (which they are in case (A.B)), then one must be simpler than the other or they are equal. Hence the third case couldn't happen.
Now I am doing case (A.C). Hints are welcome.
EDIT: I have a solution for the case $D(x)\leq\delta(y,z)$, now I just need to show that $>$ isn't possible.
EDIT: It is possible if $\delta(x,z)=\delta(y,z)$.