# Boolean Reductions problem

1. Dec 11, 2004

### DethRose

Ive been working on this problem for hours and its due on monday i really need help !!!!!!!!!

heres the boolean reduction i have to do
$$(ab)$$$$($$$$\overline c$$$$+$$$$\overline d$$$$)$$$$+$$$$(a+b)$$$$($$$$\overline c$$$$\overline d$$$$)$$

heres what ive done
***_******_ _**_ _
AB C+ABD+AC D+BC D
***_***_ _
AB(C+D)+C D (A+B)

i have no idea where to go from there...thanks for you help

the stars are just to keep the bars over top of the correct letter so you know which ones are inverted

andrew

Last edited: Dec 11, 2004
2. Dec 11, 2004

### Integral

Staff Emeritus
I am a bit confused by your symbology.

I assume + equal to AND , the bar should be NOT but what is meant by AB?

look here to learn how to do this

$$AB \overline {C}$$

3. Dec 11, 2004

### Staff: Mentor

Correction:

AB = A and B
A+B = A or B.

That seems to be the archaic form according to various websites. But that is the form that I was familiar with along time ago.

Last edited: Dec 11, 2004
4. Dec 11, 2004

### DethRose

when the numbers are right beside each other they are anded together..and when there is a plus sign it means they are ord...thats the way ive been taught to represent them haha...the bar on top means the input is inverted

5. Dec 11, 2004

### DethRose

heres the revised formula haha

$$(ab)$$$$($$$$\overline c$$$$+$$$$\overline d$$$$)$$$$+$$$$(a+b)$$$$($$$$\overline c$$$$\overline d$$$$)$$

6. Dec 12, 2004

### ehild

$$\overline{C} + \overline{D}=\overline{CD}$$

$$(A + B)\overline{CD} + (AB)\overline {CD} = \overline{CD}(AB + A +B )$$

$$A + B + AB = A + B$$

$$AB(\overline{C} + \overline{D}) + (A+B)(\overline{C}\overline{D}) = (A+B)\overline{CD}$$

ehild

7. Dec 12, 2004

### DethRose

that doesnt work because the cd in brackets doesnt have a whole bar on top of it...it is split in half

8. Dec 12, 2004

### Staff: Mentor

Correctly written, original problem should be:
$$AB(\overline{C} + \overline{D}) + (A+B)(\overline{C}\,\overline{D})$$

What if one lets:

$$W=AB$$

$$X=\overline{C}+\overline{D}$$

$$Y=A+B$$

$$Z=\overline{C}\,\overline{D}$$

Then the above original expression becomes

(WX) + (YZ)

((WX)+Y)((WX)+Z)

(W+Y)(X+Y)(W+Z)(X+Z)

or

$$\Big({(AB)+(A+B)}\Big) \Big({\overline{C}+\overline{D}+A+B}\Big)\Big({AB\overline{C}\,\overline{D}}\Big)\Big(({\overline{C}+\overline{D})\overline{C}\,\overline{D}}\Big)$$

then make use of commutative and associative properties and

a(a+b) = a+(ab) = a

so for example

(AB)+(A+B) = ((AB)+A)+B = A+B

You'll need to do a sanity check on the intermedate relationships (W, X, Y, Z). It's been awhile since I have done this.

9. Dec 12, 2004

### DethRose

i have no idea what you are doing there...there has to be a way to do it without having to replace the variables and just using the boolean rules i think