# Proving the Demorgan Law

1. Jan 28, 2009

### tomboi03

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

I tried this..
={x: x$$\in$$A, x$$\notin$$(B$$\cup$$C)}
={x: x$$\in$$A, x$$\notin$$B OR x$$\notin$$C}
={x: x$$\in$$A, x$$\notin$$B AND x$$\in$$A, x$$\notin$$C}
=(A-B)$$\cap$$(A-C)

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

={x: x$$\in$$A, x$$\notin$$(B$$\cap$$C)}
={x: x$$\in$$A, x$$\notin$$B AND x$$\notin$$C}
={x: x$$\in$$A, x$$\notin$$B OR x$$\in$$A, x$$\notin$$C}
=(A-B)$$\cup$$(A-C)

Is this wrong? What am I doing wrong?

Thank You!

2. Jan 28, 2009

### mXSCNT

There's your problem. x$$\notin$$(B$$\cup$$C) if and only if x$$\notin$$B AND x$$\notin$$C. It helps to imagine the sets as venn diagrams. Alternatively, look at it as $$\neg$$ (x $$\in$$ (B$$\cup$$C)) which becomes $$\neg$$(x $$\in$$B OR x$$\in$$C) and then the not distributes by de morgan's law for logic, producing x$$\notin$$B AND x$$\notin$$C. 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.

3. Jan 28, 2009

### 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)