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)?

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- October 15th 2008, 06:56 AMmathfanaticDeMorgan'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)? - October 15th 2008, 07:18 AMpoutsos.B
- October 16th 2008, 04:10 AMtester85
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' ) - October 16th 2008, 05:38 AMmathfanatic
- October 16th 2008, 05:44 AMtester85
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) - October 16th 2008, 06:42 AMmathfanatic
- October 16th 2008, 07:07 AMtester85
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 - October 16th 2008, 07:21 AMmathfanatic

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! - October 16th 2008, 11:49 AMpoutsos.B
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 - October 16th 2008, 12:30 PMmathfanatic

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)?