# Simplifying a term

• Oct 23rd 2009, 02:23 AM
Simplifying a term
Hello,

Can anyone help me simplifying this term?

a AND (a xor b)

I know that I can write it like this:

(a AND not b) OR (not a AND b)

But how to simplify further? Would be grateful for any help. Thanks!
• Oct 23rd 2009, 06:21 AM
PiperAlpha167
Quote:

Hello,

Can anyone help me simplifying this term?

a AND (a xor b)

I know that I can write it like this:

(a AND not b) OR (not a AND b)

But how to simplify further? Would be grateful for any help. Thanks!

(a AND not b) OR (not a AND b)

This is just equivalent to (a xor b), not the initial formula.
But, it turns out that all you need is the left disjunct.
That's the (perfect) disjunctive normal form for the initial formula.
• Oct 23rd 2009, 12:47 PM
Soroban

I'll let you supply some of the reasons . . .

Quote:

Simplify: .a AND (a XOR b)

. . $\begin{array}{ccc}a \cap \bigg[(a \cap b') \cup (a' \cap b)\bigg] & & \text{Given} \\ \\ \bigg[ a\cap (a \cap b')\bigg] \cup \bigg[a \cap(a'\cap b)\bigg] && \text{Distributive} \\ \\ \bigg[(a\cap a) \cap b'\bigg] \cup \bigg[(a\cap a') \cap b\bigg] & & \text{Associative}\end{array}$

. . . . . $\begin{array}{c}(a\cap b') \cup (f \cap b) \\ \\(a \cap b') \cup f \\ \\ a \cap b' \end{array}$

• Oct 23rd 2009, 02:10 PM
Awesome, much appreciated, thank you!
Just one question:

$
a\cap a'
$

Thats = 0 isn't it? Why did you replace it with f? And at the end you got

$
a\cap b' \cup f
$

Why is it possible to ignore the f and say that

$
(a\cap b')
$

is the result?