Logical form of "All present Kings of France are bald." I'm reading Severin Schroeder's Wittgenstein right now. In the "Between Vienna and Cambridge" chapter he introduces Russell's and Whitehead's Logical Formalization: x and y is $x \supset y$ and if x then y is $x . y$ and so on.
Now in the "Tractatus Logico-Philosophicus" chapter Schroeder writes the following:

(K1) There exists at least one present King of France.
  (K2) There exists at the most one present King of France.
  (K3) All present Kings of France are bald.  
In logical noation (Fx: x is a present King of France; Gx: x is bald):
(K1') ($\exists$x)Fx
  (K2') (x)(y)((Fx.Fy)$\supset$x=y)  
Read: Take anything x and anything y: if x is a present King of France
  and y is a present King of France, then x and y are the same (person).
  In other words, there aren't two different present kings of France.

The last paragraph puzzles me, for this really is not what I am reading there at all. Rather it'd be: 

Take anything x and anything y: if x is a present King of France THEN y is a present King of France, AND x and y are the same.

May someone please riddle me this?
 A: The formula :

(K2') $\quad \forall x \forall y ((Fx.Fy) \supset x = y)$

must be read as : take any $x$ and any $y$, if $x$ is a present King of France and $y$ is a present King of France, then $x$ and $y$ are the same thing. In other words, there aren't two different present kings of France.
The quantifier are in front of the sub-formula betwee parentheses: so their scope is all the sub-formula and the connective $\supset$ is inside the parentheses.
Your reading : take any $x$ and any $y$, if $x$ is a present King of France, then $y$ is a present King of France, and $x$ and $y$ are the same, correspond more exactly to :

$\forall x \forall y (Fx \supset Fy).x = y))$

and the two are not the same.
The first formula is equivalent to :

$\forall x \forall y (Fx \supset (Fy \supset x = y))$.

You can check with truth-tables (a contribution to mathematical logic mainly from Wittegenstein) that :

$(p.q) \supset r$ and $p \supset (q \supset r)$ 

are equivalent, but they are not equivalent to :

$(p \supset q).r$

