Proof that $x+(-x)=0$ for surreal numbers This is from Conway's on numbers and games:
$x+(-x)=0$. We have to show $x+(-x)\geq 0$ and $x+(-x )\leq 0$.  If say $(x+(-x))\ngeq 0$, we should have some $(x+(-x))^R\leq 0$, that is $x^R+(-x)\leq 0$ or $x+(-x)^L\leq 0$. But these are false, since we have by induction $x^R+(-x)^R\geq 0$, $x^L+(-x)^L\geq 0$.
I know that $x^R+(-x)\leq 0$ is a right option of $(x+(-x))$ so if its negative the number is. I don't understand where the "or $x+(-x)^L\leq 0$" come from.
And why do we have $x^R+(-x)^R\geq 0$, $x^L+(-x)^L\geq 0$ by induction?
 A: Where does $x+(-x)^L\le0$ come from?
It is a typo that your question has that my copy of the book does not, possibly related to confusion about notational conventions.
My copy says $x+-x^L$, not $x+(-x)^L$. This is because the right options of $x+(-x)$, that is, the ones denoted by "$(x+\,-x)^R$" in the book, have either the form $x^R+(-x)$ or the form $x+(-x)^R$. And by the definition of $-x$, something like $(-x)^R$ is like $-(x^L)$. By the conventions related to order of operations in regular arithmetic (since exponentiation has higher precedence than multiplication by $-1$), we write $-x^L$ to mean $-(x^L)$ and not $(-x)^L$.
Why do we have things by induction?
This is related to the paragraph just before PROPERTIES OF ORDER AND EQUALITY.

...note that almost all our proofs are inductive...we feel free to suppress all references to these inductive hypotheses

In this case, the implied inductive assumption is that $y+\,-y\ge0$ is true for all of the options of $x$, so that $x^R+\,-x^R\ge0$ and $x^L+\,-x^L\ge0$.
Where's the rest of the proof?
In case it wasn't clear, when the book says "say, $x+\,-x\ngeq0$", it's implying that we're only proving $x+\,-x\ge0$ and that the proof of $\le0$ is analogous.
