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 <====> \forall a\forall b[~(a,b)εΑxB] <=====>



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



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


\forall a( ~aεA) v \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.


Thanks