Is De Morgan's Law Always True?

[johncena]

In my textbook, the proof for demorgan's law,
(AintersectionB)* = A*unionB*
[*=complement]
starts with,saying that for all x belongs to (AintersectionB)* , x is not a member of AunionB.
But how can we say that, for example,
if A = {1,2,3} and B = (2,3,4,5} and U = {1,2,3,4,5}
(AintersectionB)^ = {1,4,5}
and AunionB = {1,2,3,4,5}
here all x which belongs to (AintersectionB)* are members of the set AunionB.

 

[statdad]

In my textbook, the proof for demorgan's law,
(AintersectionB)* = A*unionB*
[*=complement]
starts with,saying that for all x belongs to (AintersectionB)* , x is not a member of AunionB.

Are you sure it doesn't say

<br /> x \in (A \cap B)^C \Rightarrow x \not \in A \cap B<br />

because if it is written as you say, it isn't correct and must be a typographical error.

 

[JonF]

de morgan's law says: A^c U B^c = (A \cap B)^c

taking your examples:
if A = {1,2,3} and B = {2,3,4,5} and U = {1,2,3,4,5}

A^c = {4,5}
B^c = {1}

A^c U B^c = {1,4,5}
(A \cap B)={2,3}
(A \cap B)^c={1,4,5}

Now to show that de morgan's crule is true in general,
First assume x \in A^c U B^c then show x \in(A \cap B)^c

Then assume x \in(A \cap B)^cthen show x \in A^c U B^c