Given numeric forms $x =\{ X_L \,|\, X_R\}$ and $y =\{ Y_L \,|\, Y_R\}$ of two surreal numbers we say that $x \le y$ if and only if
- There is no $x_L \in X_L$ such that $y \le x_L$ and
- There is no $y_R \in Y_R$ such that $y_R \le x$.
According to this definition it seems to me that the comparison $$ \{ \,| \,2 \, \} \le \{ \, -1 \, | \, \}, $$
which is a short for $\{\varnothing\,| \,\{2\} \} \le \{ \{-1\} \, | \, \varnothing \}$, should be automatically true, since $X_L = \varnothing$ and $Y_R = \varnothing$. However, as far as I understand the surreal numbers, the expressions $ \{ \,| \,2 \, \}$ and $ \{ \, -1 \, | \, \}$ are valid numeric forms that represent numbers $1$ and $0$ respectively. Combining these two notions we arrive at $$ 1 \le 0, $$ which seems to be absurd, since one expect surreal numbers to be compatible with the natural ordering.
Obviously, I am overlooking something. I would be very grateful if someone could point out a mistake.
Both $\{|2\}$ and $\{-1|\}$ are actually equal to $0$, so your inequality is the unsurprising $0\le0$.
For reference, I'll show that $0\le\{-1|\}$. Both conditions "there is no $x_L$..." and "there is no $y_R$..." are vacuously satisfied. $\{|2\}\le0$ is analogous.
You may have confused things with $-2=\{|-1\}$ and $3=\{2|\}$. As we've just seen, it makes a big difference which set is empty.