Math Proof A-(B∩C)=(A-B)∪(A-C).

**goalkeeper00** · April 25th 2012, 03:05 PM

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).

**ignite** · April 25th 2012, 03:11 PM

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)

**Plato** · April 25th 2012, 03:17 PM

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).

$\begin{align*}A\setminus (B\cap C)&=A\cap (B\cap C)^c \\&=A\cap (B^c\cup C^c)\\&=(A\cap B^c)\cup(A\cap C^c)\\&=~?\end{align*}$

**HallsofIvy** · April 27th 2012, 07:46 AM

Or use the basic definitions: if x is in $A-(B\cap C)$ then x is in A but not in $B\cap C$ . 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 $A\cap B$ .
case 2: x is in A but not in C. Then it is in $A\cap C$ .

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