How Can I Correctly Prove the Demorgan Laws?

[tomboi03]

1st Demorgan Law
A-(B\cupC) = (A-B)\cap(A-C)

I tried this..
={x: x\inA, x\notin(B\cupC)}
={x: x\inA, x\notinB OR x\notinC}
={x: x\inA, x\notinB AND x\inA, x\notinC}
=(A-B)\cap(A-C)

2nd Demorgan Law
A-(B\capC) = (A-B)\cup(A-C)

={x: x\inA, x\notin(B\capC)}
={x: x\inA, x\notinB AND x\notinC}
={x: x\inA, x\notinB OR x\inA, x\notinC}
=(A-B)\cup(A-C)

Is this wrong? What am I doing wrong?
Please help me out!

Thank You!

 

[mXSCNT]

1st Demorgan Law
A-(B\cupC) = (A-B)\cap(A-C)

I tried this..
={x: x\inA, x\notin(B\cupC)}
={x: x\inA, x\notinB OR x\notinC}

There's your problem. x\notin(B\cupC) if and only if x\notinB AND x\notinC. It helps to imagine the sets as venn diagrams. Alternatively, look at it as \neg (x \in (B\cupC)) which becomes \neg(x \inB OR x\inC) and then the not distributes by de morgan's law for logic, producing x\notinB AND x\notinC. I assume you're allowed to use his logic rules to prove that they hold for sets. You make a similar mistake in the second one.

 

[poutsos.A]

Yes i agree with mXSCNT he explained very well where you made a mistake .

Alternatively you can prove the above by using the concept of subsets:

X=Y iff( X is a subset of Y and Y is a subset of X) iff ( xεX <===>xεY)