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**alexandros** proof:

Let :$\displaystyle \vee$ denote "or" ,and

let :$\displaystyle \wedge$ denote "and"

1)Let (x,y)εf =>

$\displaystyle [(x,y)=(1,a)\vee (x,y)=(2,b)]\Longrightarrow [(x,y)=(1,a)\vee (x,y)=(1,b)\vee (x,y) = (2,b)\vee (x,y)=(2,a)]\Longrightarrow (x,y)\ A\times B\Longrightarrow f\subset A\times B$

2) Let xεΑ => χ=1v x=2.

For x=1

put $\displaystyle y = a\Longrightarrow [y = a\vee y = b]\Longrightarrow y\in B$

Also

$\displaystyle (x,y) = (1,a)\Longrightarrow [(x,y) = (1,a)\vee (x,y) = (2,b)]\Longrightarrow (x,y)\in f$

Thus there exists y such that,$\displaystyle y\in B$ and $\displaystyle (x,y)\in f$

So much for the existence part.

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

for all ,y, z:$\displaystyle [( y\in B\wedge (x,y)\in f\wedge z\in B\wedge (x,z)\in f)\Longrightarrow y = z]$

So let :

$\displaystyle (x,y)\in f\wedge(x,z)\in f\Longrightarrow[(x,y) = (1,a)\vee (x,y) = (2,b)]\wedge[(x,z) = (1,a)\vee(x,z) = (2,b)]$

BUT $\displaystyle 1\neq 2$ (an assumption needed in the initial hypothesis for f to be a function),thus $\displaystyle 1\neq 2\Longrightarrow 1\neq 2\vee y\neq b\Longrightarrow (1,y)\neq (2,b)\Longrightarrow (x,y)\neq (2,b)$

,since x=1.

In the same way .$\displaystyle (x,z)\neq (2,b)$

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

So:

for x= 1 there exists a unique $\displaystyle y\in B$ ,such that $\displaystyle (x,y)\in f$

in the same way:

for x=2 there exists a unique $\displaystyle y\in B$ ,such that $\displaystyle (x,y)\in f$

Therefor.:

For all ,x:$\displaystyle x\in A$ there exists a unique $\displaystyle y\in B$ ,such that $\displaystyle (x,y)\in f$ .

Thus :$\displaystyle f:A\to B$