# sets prove

• Jun 3rd 2009, 01:31 AM
thereddevils
sets prove
Show that for any sets A and B ,

(AnB)'-(A'nB)=B'

Thanks for helping ..
• Jun 3rd 2009, 03:01 AM
Moo
Hello,
Quote:

Originally Posted by thereddevils
Show that for any sets A and B ,

(AnB)'-(A'nB)=B'

Thanks for helping ..

For two sets P and Q, $P-Q=P\cap Q'$

Thus $(A\cap B)'-(A'\cap B)=(A\cap B)'\cap (A'\cap B)'$

By de Moivre's formula, this is :
$[(A\cap B)\cup (A'\cap B)]'$

And maybe it's clear for you that $(A\cap B)\cup (A'\cap B)=B$ ?

If you don't see it :
- draw a diagram
- or expand, and use the fact that $A\cup A'=E$, where E is the universe.
• Jun 3rd 2009, 03:07 AM
Plato
Quote:

Originally Posted by thereddevils
Show that for any sets A and B ,
(AnB)'-(A'nB)=B'

$\left( {A \cap B} \right)^c \backslash \left( {A^c \cap B} \right) = \left( {A \cap B} \right)^c \cap \left( {A^c \cap B} \right)^c = \left( {A^c \cup B^c } \right) \cap \left( {A \cup B^c } \right)$
• Jun 3rd 2009, 01:58 PM
Soroban
Hello, thereddevils!

Quote:

Show that for any sets A and B: . $
(A \cap B)'-(A' \cap B) \:=\:B'$

I'll signify a couple of axioms:

. . $(1)\;\;P \cap P' \:=\: \emptyset \qquad (2)\;\; P \cup \emptyset \:=\: P$

$\begin{array}{ccccc}
(A \cap B)' - (A' \cap B) & = & (A \cap B)' \cap (A' \cap B)' && \text{d{e}f. Subtraction} \\ \\

& = & (A' \cup B') \cap (A \cup B') && \text{DeMorgan} \\ \\

& = & (A' \cap A) \cup B' && \text{Distributive} \\ \\

& = & \emptyset \cup B' && \text{axiom (1)} \\ \\

& = & B' && \text{axiom (2)} \end{array}$