sets prove

**thereddevils** · June 3rd 2009, 01:31 AM

Show that for any sets A and B ,

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

Thanks for helping ..

**Moo** · June 3rd 2009, 03:01 AM

Hello,

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.

**Plato** · June 3rd 2009, 03:07 AM

Originally Posted by thereddevils

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

$\left( {A \cap B} \right)^c \backslash \left( {A^c \cap B} \right) = \left( {A \cap B} \right)^c \cap \left( {A^c \cap B} \right)^c = \left( {A^c \cup B^c } \right) \cap \left( {A \cup B^c } \right)$

**Soroban** · June 3rd 2009, 01:58 PM

Hello, thereddevils!

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

I'll signify a couple of axioms:

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

$\begin{array}{ccccc} (A \cap B)' - (A' \cap B) & = & (A \cap B)' \cap (A' \cap B)' && \text{d{e}f. Subtraction} \\ \\ & = & (A' \cup B') \cap (A \cup B') && \text{DeMorgan} \\ \\ & = & (A' \cap A) \cup B' && \text{Distributive} \\ \\ & = & \emptyset \cup B' && \text{axiom (1)} \\ \\ & = & B' && \text{axiom (2)} \end{array}$

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