proof:

Let :

denote "or" ,and

let :

denote "and"

1)Let (x,y)εf =>

2) Let xεΑ => χ=1v x=2.

For x=1

put

Also

Thus there exists y such that,

and

So much for the existence part.

Now for the uniqueness part.For that we must prove:

for all ,y, z:

So let :

BUT

(an assumption needed in the initial hypothesis for f to be a function),thus

,since x=1.

In the same way .

Hence (x,y)=(1,a) and (x,z) =(1,a) and so y=z

So:

for x= 1 there exists a unique

,such that

in the same way:

for x=2 there exists a unique

,such that

Therefor.:

For all ,x:

there exists a unique

,such that

.

Thus :