# DeMorgan's Law?

• Oct 15th 2008, 07:56 AM
mathfanatic
DeMorgan's Law?
I am not sure if I am correct, but would:

(A U C) n (B U C) be the same as:

(A n C) U (B n C)?
• Oct 15th 2008, 08:18 AM
poutsos.B
Quote:

Originally Posted by mathfanatic
I am not sure if I am correct, but would:

(A U C) n (B U C) be the same as:

(A n C) U (B n C)?

NO

CU(A^B) = (CUA)^(CUB) and

C^(AUB) = (C^A)U(C^B)
• Oct 16th 2008, 05:10 AM
tester85
If you apply demorgan theorem, you will get the following equation listed below :

((A U C) n (B U C))'' = ( A U C )' U ( B U C )'
= ( A' n C' ) n ( B' n C' )
• Oct 16th 2008, 06:38 AM
mathfanatic
Quote:

Originally Posted by tester85
If you apply demorgan theorem, you will get the following equation listed below :

((A U C) n (B U C))'' = ( A U C )' U ( B U C )'
= ( A' n C' ) n ( B' n C' )

Thanks for the reply and the help. I am just a tad bit confused!

How did you end up going from ( A U C )' U ( B U C )'
= ( A' n C' ) n ( B' n C' )?

I try and try but just do not understand DeMorgan's Law at all (Headbang)
• Oct 16th 2008, 06:44 AM
tester85
Sorry typo. Should be :

((A U C) n (B U C))'' = (( A U C )' U ( B U C )')'
= (( A' n C' ) n ( B' n C' ))'

It is done based on the following formulas :

( A U B )' = A' n B'

( A n B )' = A' U B'

Sorry for any inconvenience caused. Learning along the way.(Clapping)
• Oct 16th 2008, 07:42 AM
mathfanatic
Quote:

Originally Posted by tester85
Sorry typo. Should be :

((A U C) n (B U C))'' = (( A U C )' U ( B U C )')'
= (( A' n C' ) n ( B' n C' ))'

It is done based on the following formulas :

( A U B )' = A' n B'

( A n B )' = A' U B'

Sorry for any inconvenience caused. Learning along the way.(Clapping)

No inconvenience by you, it's just me and my stubborn head that can't wrap my brain around this!!

So do you have to first change the symbols to all to union in order to apply the law and chang to intersection?
• Oct 16th 2008, 08:07 AM
tester85
Hmmm. How should i put it in words. Maybe you would like to look in the link provided below for further clarification.

http://en.wikipedia.org/wiki/De_Morgan_duality
• Oct 16th 2008, 08:21 AM
mathfanatic
Quote:

Originally Posted by tester85
Hmmm. How should i put it in words. Maybe you would like to look in the link provided below for further clarification.

De Morgan's laws - Wikipedia, the free encyclopedia

Believe it or not, I have this up in my other window and have been looking at it for over an hour now! (Headbang)

Maybe if I explain what is ging on in my brain that would help!

I am thinking that:
(A U C) n (B U C) would then be (A n C) U (B n C)

But you said that (A U C) n (B U C) becomes (A U C) U (B U C) which is then (A n C) n (B n C).

So how come it didn't go from (A U C) n (B U C) to (A n C) U (B n C)? That is where I am confused. Sorry to be such a pain-I really do appreciate your help and patience!
• Oct 16th 2008, 12:49 PM
poutsos.B
Quote:

Originally Posted by mathfanatic
I am not sure if I am correct, but would:

(A U C) n (B U C) be the same as:

(A n C) U (B n C)?

Let C={1,2,3}..........B = {1,a,b}.........A = { 1,a,3} and

AUC = { 1,a,3}U{1,2,3}= { 1,2,3,a}.......................................... ..........1

BUC = {1,a,b}U{1,2,3}={1,2,3,a,b}....................... ..............................2

Hence: (AUC)n(BUC) = {1,2,3,a}U{1,2,3,a,b} = {1,2,3,a,b}....................................... ...............3

But :

AnC = { 1,a,3}n{1,2,3} = {1,3}............................................. ...................4

BnC = {1,a,b}n{1,2,3} = { 1 }................................................. ...............5

Hence: (A n C) U (B n C) = {1,3}U{1 } = { 1,3}.............................................. ....................6

THUS:

(A U C) n (B U C) IT IS NOT the same as:

(A n C) U (B n C)

Note to show that a particular identity does not hold one counter example is enough ,although in our case more than one example could show that.

The identities in concern are :

CU(AnB) = (CUA)n(CUB) .................................................. ..........a

...................................and............ .................................................

Cn(AUB) = (CnA)U(CnB)....................................... ...........................b

In a U distributes over n and in b n distributes over U.

Another way to check if a=b would be if the DOUBLE implication ,

.............................xεCU(AnB)<=====>xεCn( AUB)............................................

......................................holds....... .................................................. ..............

since :CU(AnB) = (CUA)n(CUB) AND Cn(AUB) = (CnA)U(CnB)....................................... ................................................

De Morgan has nothing to do with the above
• Oct 16th 2008, 01:30 PM
mathfanatic
Quote:

Originally Posted by mathfanatic
No inconvenience by you, it's just me and my stubborn head that can't wrap my brain around this!!

So do you have to first change the symbols to all to union in order to apply the law and chang to intersection?

Ooh wait, I think I finally figured it out after staring at this all afternoon!!!!!

(Wait)
I think I see how you got that (AUC)=(AnC) and (BUC)=(BnC) but I am still confused on why the n in the original problem doesn't just go to u so that you would have (AnC) u (BnC) instead of (AnC) n (BnC)?