Boolean algebra help please!

[coach94]

I can't use K-Map's i only can use boolean algebra to solve these problems

This is an example of one of the many things i need to simplify using boolean algebra

D'C'B'A+D'C'BA'+D'C'BA+D'CBA'+D'CBA+DC'B'A'+DC'BA

Any kind of help is appreciated, I honestly dont even know where to start.

Edited September 17, 2013 by coach94

[studiot]

Do you know the rules of boolean algebra as algebra?

 

A+AB = A

A(A+B)=A

A(A'+B)=AB

A+A'B=A+B

A+BC=(A+B)(A+C)

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

A'B'=(A+B)'

 

The last two make up De Morgan's Theorem.

 

Alternatively you could use Venn Diagrams.or switch diagrams

Edited September 17, 2013 by studiot

[coach94]

Do you know the rules of boolean algebra as algebra?

 

A+AB = A

A(A+B)=A

A(A'+B)=AB

A+A'B=A+B

A+BC=(A+B)(A+C)

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

A'B'=(A+B)'

 

The last two make up De Morgan's Theorem.

 

Alternatively you could use Venn Diagrams.or switch diagrams

I've seen those rules a few times before, But i dont see how they will help me when I have 4 variables not just 2 or 3?

[studiot]

Come on, can you not start collecting terms together and simplifying?

 

You also have a bunch of more basic rules like

 

A.A=A

A+A=A

A.1=A

A+A'=1

AA'=0

(A')'=A

A+B=B+A

A+B+C=A+(B+C)=(A+B)+C

ABC=A(BC)=(AB)C

A(B+C)=AB+AC

 

So for instance in your example you have

 

D'C'BA+D'C'BA'

 

=D'C'B(A+A')

 

and what is (A+A')?

Edited September 17, 2013 by studiot

[coach94]

Come on, can you not start collecting terms together and simplifying?

 

You also have a bunch of more basic rules like

 

A.A=A

A+A=A

A.1=A

A+A'=1

AA'=0

(A')'=A

A+B=B+A

A+B+C=A+(B+C)=(A+B)+C

ABC=A(BC)=(AB)C

A(B+C)=AB+AC

 

So for instance in your example you have

 

D'C'BA+D'C'BA'

 

=D'C'B(A+A')

 

and what is (A+A')?

(A+A') would be 1 right?

[EdEarl]

Just so there is no misunderstanding,

 

1. Does A+B mean A or B or A exclusive or B?

2. Does D' mean not D?

3. Does AB mean A and B?

4. Is there any significance to your underlining the entire equation: D'C'B'A+D'C'BA'+D'C'BA+D'CBA'+D'CBA+DC'B'A'+DC'BA? If the underline means the entire equation is not equation then use one of the laws to change it to another form without the not. If this is why you cannot start, make the whole thing look simpler by renaming each of the term+term+term....

 

I recommend you put parenthesis around a couple of the terms and see if you can simplify the things in parenthesis. Use the laws (monotone, nonmonotone, and De Morgan's) to simplify.

 

Try something and show your work. Check your work with a K-map, just don't turn it in.

 

I'll let you do this studiot.

Edited September 17, 2013 by EdEarl

[studiot]

Yes indeed.

 

So does that not simplyfy the second and third terms combined?

 

You need to play with this idea and see how far you can simplify things this way.

 

I will be away for about half an hour now so happy playing.

[coach94]

Just so there is no misunderstanding,

 

1. Does A+B mean A or B or A exclusive or B?

2. Does D' mean not D?

3. Does AB mean A and B?

4. Is there any significance to your underlining the entire equation: D'C'B'A+D'C'BA'+D'C'BA+D'CBA'+D'CBA+DC'B'A'+DC'BA? If the underline means the entire equation is not equation then use one of the laws to change it to another form without the not. If this is why you cannot start, make the whole thing look simpler by renaming each of the term+term+term....

 

I recommend you put parenthesis around a couple of the terms and see if you can simplify the things in parenthesis. Use the laws (monotone, nonmonotone, and De Morgan's) to simplify.

 

Try something and show your work. Check your work with a K-map, just don't turn it in.

1. means or

2. correct

3. yes

4. no just underlined it just because ( probably shouldn't do that )

 

Now when Im doing this, can i take say an A' from the second set of terms and match it with one from lets say the second to last set of terms? Do i have to go in a specific order? And do the rules apply to other variables the same? or do they work just as a or b?

[studiot]

 

Now when Im doing this, can i take say an A' from the second set of terms and match it with one from lets say the second to last set of terms? Do i have to go in a specific order? And do the rules apply to other variables the same? or do they work just as a or b?

 

 

A,B, C, D, E, F , elephant, monkey, tangerine are all dummy variables the same rules apply to them all. So where I have written A you could write any other variable and the same for B or C etc.

 

The only restrictions to order are as I outlined in my list of rules. So yes, take the first and last if you like, though I recommend seeing what you can make of the last two terms. Can you simplify them together?

Edited September 17, 2013 by studiot

[coach94]

So what I got from the last two would be

 

DC'B'A' + DC'BA =

 

DC'B' + DC'B =

 

DC' + DC' =

 

C'(D+C)?

Edited September 17, 2013 by coach94

[studiot]

 

DC'B'A' + DC'BA =

 

DC'B' + DC'B =

 

The second does not follow from the first, since you have one A' and one A

 

How about trying the fourth and fifth terms?

Edited September 17, 2013 by studiot

[coach94]

 

The second does not follow from the first, since you have one A' and one A

 

How about trying the fourth and fifth terms?

but wouldnt

A+A'=1

imply that those a's would be 1?

and for the 4th and 5th terms

it would go

 

D'CB(A+A')?

 

which would be

 

D'CB?

[studiot]

D'CB is correct.

 

You are getting there.

 

Now what do you see if you add what you got from simplyfying the second and third terms to what you got in post 12?

 

D'C'B + D'CB

 

Can you take the next step with these pairs?

ie which variable can you eliminate next?

 

You will then have reduced four of the seven terms to 1.

Not bad going.

 

But I must leave you to work on the others a bit we do not do all your work for you here.

Edited September 17, 2013 by studiot

[coach94]

would you be able to check if im going in the right direction with this? Just want to make sure I got the first one done correctly i have about 30 more to do lol.

 

from the whole example i provided, i got

 

B+D'C'+D'+C'

would you be able to check if im going in the right direction with this? Just want to make sure I got the first one done correctly i have about 30 more to do lol.

 

from the whole example i provided, i got

 

B+D'C'+D'+C'

actually cant you combine it to be

 

B+D'+C'

[studiot]

How to perform a quick check of your work?

 

Well boolean variables can take on only two values viz 0 or 1.

 

So assign values to A, B, C, D ...etc and substitute

 

In your case let A=B=C=D=1

 

your expression comes to

 

0+0+0+0+0+0+0=0

 

or let A=B=C=D=0

 

your expression comes to

 

 

0+0+0+0+0+0+0=0

 

So any equivalent expression must evaluate to the same.

 

However this check is like an accountant balancing the books, it is not perfect.

That is if the expressions disagree there is definitely a mistake somewhere, but if they agree there is no guarantee that you have not made self cancelling mistakes.

Edited September 18, 2013 by studiot

[Delbert]

I've only got as far as trying to decide the meaning of the symbols!

 

Would I be right in assuming:-

' = not

+ = or

. = and

 

And I was trying to draw a Venn diagram for four items, but failed.

[EdEarl]

Would I be right in assuming:-

' = not

+ = or

. = and

 

Yes

 

And I was trying to draw a Venn diagram for four items, but failed.

Karnaugh maps are commonly used to help with Boolean equations.