What is $(A\setminus B)^c$? It seems to me that it is equal to $B\cup A^c$, isn't it? Please help me to verify if there is a mistake?
Complement of the difference of the sets
0
$\begingroup$
elementary-set-theory
-
0@Martin: that's certainly right. I never used the truth tables though. Maybe I will start now. – 2011-07-06
2 Answers
4
Using the property $A \backslash B = A \cap B^c$ and the De-Morgan's laws you get the following chain of equalities:
$ (A \backslash B)^c = (A \cap B^c)^c = A^c \cup (B^c)^c = A^c \cup B $
2
Yes, $ (A - B)^c = (A \cap B^c )^c = A^c \cup (B^c )^c = A^c \cup B. $