# De Morgan's laws

• Dec 18th 2008, 12:37 AM
Showcase_22
De Morgan's laws
Quote:

The set (A\B)\C is equal for all sets A,B and C to one of the following sets:

i). A\(B \$\displaystyle \cap\$ C)
ii). A\(B\C)
iii). A\(B \$\displaystyle \cup\$ C)

Which one?
This question can be done using truth tables, but it would take a really long time!

I was wondering if there was a way to do it using De Morgan's laws. This is what i've done so far:

(A\B)\C=(A\B) \$\displaystyle \cap C^c\$=\$\displaystyle (A \cap B^c)\cap C^c\$

I'm not sure where to go from here. I tried expanding the other expressions to see if I could get to this, but it hasn't worked.
• Dec 18th 2008, 12:54 AM
Chop Suey
\$\displaystyle (A \cap B^c)\cap C^c = A \cap (B^c \cap C^c) = A \cap (B \cup C)^c\$

Looks familiar, doesn't it?
• Dec 18th 2008, 01:12 AM
Showcase_22
\$\displaystyle
(A \cap B^c)\cap C^c = A \cap (B^c \cap C^c) = A \cap (B \cup C)^c
\$

lol, it does (and is):

\$\displaystyle
(A \cap B^c)\cap C^c = A \cap (B^c \cap C^c) = A \cap (B \cup C)^c
\$=A\(B \$\displaystyle \cup\$ C)

So when you have an expression with all \$\displaystyle \cup\$'s or \$\displaystyle \cap\$'s you can change around the brackets? (ie. It's associative?)
• Dec 18th 2008, 01:16 AM
Chop Suey
Quote:

Originally Posted by Showcase_22
So when you have an expression with all \$\displaystyle \cup\$'s or \$\displaystyle \cap\$'s you can change around the brackets? (ie. It's associative?)

Yes, it's associative.