0
$\begingroup$

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?

  • 0
    @Martin: that's certainly right. I never used the truth tables though. Maybe I will start now.2011-07-06

2 Answers 2

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. $