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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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)?
NOQuote:
Originally Posted by mathfanatic
CU(A^B) = (CUA)^(CUB) and
C^(AUB) = (C^A)U(C^B)
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' )
Quote:
Originally Posted by tester85
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)
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!!Quote:
Originally Posted by tester85
So do you have to first change the symbols to all to union in order to apply the law and chang to intersection?
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
Quote:
Originally Posted by tester85
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!
Let C={1,2,3}..........B = {1,a,b}.........A = { 1,a,3} andQuote:
Originally Posted by mathfanatic
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
Quote:
Originally Posted by mathfanatic
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)?