Are the following proofs correct??

a) AxB =Φ <====> [ (a,b)εAxB <-----> (a,b)εΦ] <=====>[(aεA & bεB)<------>(aεΦ & bεΦ)] <====>[(aεA----->aεΦ) & (bεB------>bεΦ)] ======>(aεA----->aεΦ)====> A= Φ ====> A=Φ v B=Φ.

Ηence:..... AxB =Φ =====> A=Φ v B=Φ

b) AxB =Φ <====> $\displaystyle \forall a\forall b$[~(a,b)εΑxB] <=====>

$\displaystyle \forall a\forall b$ [ ~aεA v ~bεB] <========>

$\displaystyle \forall a\forall b$( ~aεA) v $\displaystyle \forall a\forall b$(~bεB) <======>

$\displaystyle \forall a$( ~aεA) v $\displaystyle \forall b$(~bεB) <======>

A= Φ v B=Φ ........justify your answer,if possible

Where:.........AxB ={ (a,b): aεA & bεB } , Φ is the empty set and " ~" means not.