# Thread: How do I prove this..

1. ## How do I prove this..

Hi,

I have an exam coming up tomorrow and am having trouble trying to prove this theorem:

Question: Using DeMorgans Theorem, show that:

I just haven't got a clue how to tackle this one. Do I only need to use DeMorgans or are there other laws I have to use first??

2. Hello, atwarwithmaths!

We use DeMorgan's Theorem just once.
Then we use a variety of other laws:
. . $\displaystyle (1)\;A \cap \overline{A} \:=\:\emptyset,\quad (2)\;A \cup \emptyset \:=\:A,\quad\text{Distributive Laws},\quad\text{Commutative Laws}$

Using DeMorgans Theorem, show that:

$\displaystyle (A \cap \overline{B}) \cup (\overline{A} \cap B) \;=\;(A \cup B)\cap (\overline{A \cap B})$
On the right, we have:

$\displaystyle (A \cup B) \:\cap \\overline{A \cap B}) \;=\;(A \cup B) \:\cap\: (\overline{A} \cup \overline{B})\qquad\text{(DeMorgan)}$

. . . . . . . . . . . . $\displaystyle = \;\bigg[(A \cup B) \cap \overline{A}\,\bigg] \;\cup \;\bigg[(A \cup B) \cap \overline{B}\,\bigg] \qquad\text{(Distr.)}$

. . . . . . . . . . . . $\displaystyle = \;\bigg[(A \cap \overline{A}) \cup (B \cap \overline{A})\bigg] \;\cup \;\bigg[(A \cap \overline{B}) \cup (B \cap \overline{B})\bigg]\qquad\text{(Distr.)}$

. . . . . . . . . . . . $\displaystyle =\;\bigg[\,\emptyset \cup (B \cap \overline{A})\bigg] \;\cup \;\bigg[(A \cap \overline{B}) \cup \emptyset\,\bigg]\qquad\text{(Law 1)}$

. . . . . . . . . . . . $\displaystyle = \;(B \cap \overline{A}) \;\cup\A \cap \overline{B}) \qquad\text{(Law 2)}$

. . . . . . . . . . . . $\displaystyle = \;(A \cap \overline{B})\:\cup\\overline{A} \cap B)\qquad\text{(Comm.)}$