If $x\not\leq y$, then is $x>y$, or $x\geq y$? I'm currently reading about surreal numbers from here. 
At multiple points in this paper, the author has stated that if $x\not\leq y$, then $x\geq y$. 
Shoudn't the relation be "if $x\not\leq y$, then $x>y$"?
Hasn't the possibility of $x=y$ already been negated when we said $x\not\leq y$? 
Thanks in advance. 
 A: The paper defines $\leq$ and it is not necessarily the "standard" $\leq$ that you and I are used to. It is often better to substitute some other symbol like $x\preceq y$, and realize we have to prove stuff about this symbol before we can use it.
In particular, in that paper, it is proved in Theorem 1.4 that $x\not\leq y$ implies $y\leq x$. It does not prove that $y<x$. 
So trying to reason about $\leq$ as it if means "less than or equal" is a mistake. There is no definition of "less than" to use here.  There is only $\leq$.
There are examples where $x\leq y$ and $y\leq x$. This means that if we define $x<y$ in terms of $x\leq y$ and $x\neq y$ then we get $x<y$ and $y<x$ and therefore we find that $<$ defined this way is non-transitive.
I think the name "surreal" should give a strong hint at the non-standard nature of these "numbers."
A: In the paper linked to, there appears to be only a single primitive relation, written $x\leq y$, of which $y\geq x$ is just a notational variant. The equals part of the symbol is present because $x\leq x$ holds for all $x$ (theorem$~1.2$) but this is just a mnemonic device without formal implications. The negation of the relation $x\leq y$ is $x\not\leq y$, and it is written like that to remind the reader that it only affirms the absence of the relation $x\leq y$, not the presence of the relation $x\geq y$. In fact $x\not\leq y$ does imply $x\geq y$, but that is only shown in theorem$~1.4$.
Equality is not an explicitly mentioned relation (at least it seems so scanning rapidly), and so there is no reason to introduce $x>y$ as as shorthand for $x\geq y\land x\neq y$; for this reason theorem$~1.4$ is formulated the ways it is. But of course the hypothesis $y\not\leq x$ does exclude the possibility that $y$ actually is $x$; indeed the notation $x>y$ is introduced just after that theorem. However I do not agree with the given motivation "Theorem 1.4 means that the the surreal numbers are totally ordered", because they are not even partially ordered, merely preordered. In view of theorem 1.2 it is what I would call a total preorder (not sure the term is in use), a relation in which every pair is comparable at least one way, possibly both ways.
This is the next point; the paper states that $x\leq y\land x\geq y$ is not only possible when $x=y$. It introduces the relation $x\equiv y$ for $x\leq y\land x\geq y$, and given transitivity, this is an equivalence relation. What is really totally ordered is the set of equivalence classes for this equivalence relation.
A: You are correct, if we were speaking of $\leq/\geq$ relations we know and love, as standard ordering relations on the reals: The negation of $x \leq y$ is exactly $x > y$, and that would be the correct assertion if we were talking about a "trichotomous" ordering, where we take that for any two real numbers, one and only one of the following hold. $x\lt y, \lor x = y, \lor x>y$.
But your text is not wrong that $x \nleq y \implies x\geq y$, (that is, the right hand side is implied by the left hand side, and this would be a valid implication in even in the standard real numbers). And it seems your text is using strictly $\leq$ and $\geq$ so that for any two numbers $x, y$, we have one and only one of the following relations to consider: $x \geq y$ or $x \leq y$, andthese relations do not necessarily hold the same properties we know and love with respect to their standard meanings on the reals.
A: Here's another way of looking at it.
$$(x\le{y})\leftrightarrow (x<y \lor x=y)$$
$$(x\not\le{y})\leftrightarrow(x>y\land{x\ne{y}})$$
$$x\not\le{y}\rightarrow{x\ne{y}}$$
Therefore:
$$x\not\le{y}\not\rightarrow{{(x\ge{y})}}$$
