Math Proof A-(B∩C)=(A-B)∪(A-C).

• Apr 25th 2012, 03:05 PM
goalkeeper00
Math Proof A-(B∩C)=(A-B)∪(A-C).
I have no idea how to solve this. Any help would be appreciated! The statement is
For any sets A, B, and C that are subsets of a universal set U, A-(B∩C)=(A-B)∪(A-C).
• Apr 25th 2012, 03:11 PM
ignite
Re: Math Proof A-(B∩C)=(A-B)∪(A-C).
Use following properties:
X-Y=X∩Y'
(X∩Y)'=X'∪Y'
A-(B∩C)=A∩(B∩C)'=A∩(B'∪C')=(A∩B')∪(A∩C')=(A-B)∪(A-C)
• Apr 25th 2012, 03:17 PM
Plato
Re: Math Proof A-(B∩C)=(A-B)∪(A-C).
Quote:

Originally Posted by goalkeeper00
I have no idea how to solve this. Any help would be appreciated! The statement is
For any sets A, B, and C that are subsets of a universal set U, A-(B∩C)=(A-B)∪(A-C).

\displaystyle \begin{align*}A\setminus (B\cap C)&=A\cap (B\cap C)^c \\&=A\cap (B^c\cup C^c)\\&=(A\cap B^c)\cup(A\cap C^c)\\&=~?\end{align*}
• Apr 27th 2012, 07:46 AM
HallsofIvy
Re: Math Proof A-(B∩C)=(A-B)∪(A-C).
Or use the basic definitions: if x is in $\displaystyle A-(B\cap C)$ then x is in A but not in $\displaystyle B\cap C$. That, in turn, means it is either NOT in B or NOT in C.
case 1: x is in A but not in B. Then it is in $\displaystyle A\cap B$.
case 2: x is in A but not in C. Then it is in $\displaystyle A\cap C$.