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idi
 Post subject: is that proof correct
PostPosted: Mon, 31 Jan 2011 02:29:58 UTC 
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Joined: Fri, 28 May 2010 02:57:26 UTC
Posts: 53
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).

Where Φ Stands for the empty set
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