The formula that the PS3 uses to calculate our rankings is quite simple really.
Rankings =
sqrt(a
2+b
2)[a/sqrt(a
2+b
2)
+ bi/sqrt(a
2+b
2)]
= sqrt(a
2+b
2)[cos(t) + sin(t) i]
= |z|e
ti
= e
ln|z|+tiIn more modern language, the theorem can also be phrased as follows: if
- 0 ? U ? V ? R ? 0
- dim(U) + dim(R) = dim(V)
Here R plays the role of im T and U is ker T.
In the finite-dimensional case, this formulation is susceptible to a generalization: if
- 0 ? V1 ? V2 ? ... ? Vr ? 0
is an exact sequence of finite-dimensional vector spaces, then
The rank-nullity theorem for finite-dimensional vector spaces may also be formulated in terms of the index of a linear map. The index of a linear map T : V ? W, where V and W are finite-dimensional, is defined by
- index T = dim(ker T) - dim(coker T).
Intuitively, dim(ker T) is the number of independent solutions x of the equation Tx = 0, and dim(coker T) is the number of independent restrictions that have to be put on y to make Tx = y solvable. The rank-nullity theorem for finite-dimensional vector spaces is equivalent to the statement
- index T = dim(V) - dim(W).
Next, we eliminate all free variables from f by quantifying them existentially: if, say, x1...xn are free in f, we form 
. If ? is satisfiable in a structure M, then certainly so is f and if ? is refutable, then 
is provable, and then so is ¬f, thus f is refutable. We see that we can restrict f to be a sentence, that is, a formula with no free variables.
Finally, we would like, for reasons of technical convenience, that the prefix
of f (that is, the string of quantifiers at the beginning of f, which
is in normal form) begin with a universal quantifier and end with an
existential quantifier. To achieve this for a generic f (subject to
restrictions we have already proved), we take some one-place relation
symbol F unused in f, and two new variables y and z.. If f = (P)F, where (P) stands for the prefix of f and F for the matrix (the remaining, quantifier-free part of f) we form ![\psi = \forall y (P) \exists z ( \Phi \wedge [ F(y) \vee \neg F(z) ] )](http://upload.wikimedia.org/math/e/1/a/e1a4fcfcbc02df0ea5046381d435ae56.png)
. Since 
is clearly provable, it is easy to see that f = ? is provable.
Our generic formula f now is a sentence, in normal form, and its
prefix starts with a universal quantifier and ends with an existential
qu