# HW Help: Boolean Simplification

Discussion in 'Homework Help' started by rtran2015, Oct 3, 2015.

1. ### rtran2015 Thread Starter New Member

Oct 3, 2015
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I have a couple questions (#) for my hw, below is what I worked on with things to help better understand the questions

(-[_]) = "not" [_]
AB = A*B = A & B
A+B = A or B

Given: Boolean Theorems T1-T12

Simplify each Boolean Theorem

(a) Y = (-A)BC + (-A)B(-C)
(b) Y = (-A)(-B)(-C) + A(-B)
(c) Y = ABC(-D) + A(-B)(-C)(-D) + (-(A + B + C + D))

Work:
(a): = (-A)B; T5 Complements
(b): = How should I begin this process?
Using the truth tables I was able to derive

=(-A)(-B)(-C) + A(-B)(-C) + A(-B)C
= (-A)(-B)(-C) + A(-B)C = (-B); using T5
= A(-B)(-C) + A(-B)C = A(-B); using T5
=(-B) + A(-B);
Though I'm not even sure this is right
(1) How should I simplify if I didn't use the truth table?

(c): Used the same logic as (b)
= (-B)(-C)(-D) + (-D)

Some needed clarification:

(2) B(-B) = 0; T5, but is it possible with multiple variables? AB+A(-B) = A?
(3) Can we use single variable theorem (T1-T5) with multiple variables?
(4) AB + (-A)(-B)D; Can this be simplified more?

Note:

Let me know if I need to explain more on a certain area (i.e. if you need all the boolean theorems)
Give me simple equations to solve to better convey an idea
Let me also know if you prefer the work to be at the top of the page or bottom if you care
Book: Digital Design and computer Architecture by Harris and Harris, Exercise 2.14

2. ### rtran2015 Thread Starter New Member

Oct 3, 2015
13
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(-(ABCD)) = (-A)+(-B)+(-C)+(-D); via deMorgan's

(5) (-(A(-B))) = -AB; is this possible?

(6) is there a way to make it so a character can have a line above it i.e. A_var, or a cleaner way to portray "NOT"

3. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,922
601
For b I got:
y=a'b'c'+ab' distributive law
=b'(a'c'+a) t10: a+a'b=a+b
=b'(a+c') distributive law
=ab'+b'c'

4. ### WBahn Moderator

Mar 31, 2012
20,068
5,666
It's hard to know what you are and aren't allowed to do when your assignment says to use theorems T1 through T12 and you haven't told us what theorems T1 through T12 are.

5. ### WBahn Moderator

Mar 31, 2012
20,068
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What happens if you "factor out" (-B) from both terms?

Now consider the following:

XY + (-X) = XY + (-X)(Y + (-Y))
= XY + [(-X)Y] + (-X)(-Y)
= XY + [(-X)Y + (-X)Y] + (-X)(-Y)
= [XY + (-X)Y] + [(-X)Y + (-X)(-Y)]
= [X + (-X)]Y + (-X)[Y + (-Y)]
= Y + (-X)

This is merely

X = X(1) combined with 1 = (Y + (-Y))

Look at this carefully. What you are claiming is that

(-A)(-C) + AC = 1

Does that ring true?

This is just the reverse of the step before the last one.

XY + X(-Y) = X(Y + (-Y)) = X(1) = X

Assuming that this was correct to this point, does the value of A matter? What happens if you "factor out" (-B) from both terms.

Though I'm not even sure this is right
(1) How should I simplify if I didn't use the truth table?
[/QUOTE]

Using Boolean Algebra, as I've been doing.

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6. ### rtran2015 Thread Starter New Member

Oct 3, 2015
13
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for:
=b'(a'c'+a) t10: a+a'b=a+b

I understand if (a'c' +ac') = c'

How does (a'c' +a) = (a+c') through combining?
How are you removing a' but still keep a?

7. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,922
601
I told you, it is t10, go look up how t10 is derived.

8. ### rtran2015 Thread Starter New Member

Oct 3, 2015
13
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T1: Identity B*1=B; T1': B+0=B
T2: Null Element B*0=0; T2': B+1=1
T3: Idempotency BB=1; T3': B+B=1
T4: Involution B''=B
T5: Complements BB'=0; T5': B+B'=1
T6: Commute BC=CB; T6': B+C=C+B
T7: Associate (BC)D=B(CD); T7': (B+C)+D=B+(C+D)
T8: Distribute (BC)+(BD)=B(C+D); T8': (B+C)(B+D)=B+(CD)
T9: Covering B(B+C)=B; T9':B+(BC)=B
T10: Combining (BC)+(BC')=B T10': (B+C)(B+C')=B

T11: Consensus (BC)+(B'D)+(CD) T11': (B+C)(B'+D)(C+D)
=BC+B'D =(B+C)(B'+D)

T12: DeMorgan's [B[0]*B[1]*...]' T12': [B[0]+B[1]+...]'
=(B'[0]+B'[1]+...) =(B'[0]*B'[1]...)

Modrators note: added a space between : and ( to avoid conversion to smilies

Last edited by a moderator: Oct 3, 2015
9. ### rtran2015 Thread Starter New Member

Oct 3, 2015
13
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the frowny face is ': ('

10. ### shteii01 AAC Fanatic!

Feb 19, 2010
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T10a here: http://www.ee.surrey.ac.uk/Projects/Labview/boolalgebra/
The first example has the proof of T10.

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Oct 3, 2015
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12. ### rtran2015 Thread Starter New Member

Oct 3, 2015
13
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I understand now

thanks

13. ### rtran2015 Thread Starter New Member

Oct 3, 2015
13
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Given:
ABCD' + A'B'C'D';

D'(ABC +A'B'C'); T8
D'(1); T5

Is this allowed?

14. ### WBahn Moderator

Mar 31, 2012
20,068
5,666
Just think about it for a moment.

You are claiming that

ABC + A'B'C' = 1

Does this make any more sense then when I asked if

(-A)(-C) + AC = 1

rang true back in Post #5.

Can you think of at least one combination of inputs for which

ABC + A'B'C' = 0

If you can, then you KNOW that the claim you are making is not correct.

Well, what if A = 0. That guarantees that the first term, ABC, is 0 regardless of what we choose for B and C. So now is there a choice we can make for B (or C) that will make the second term, A'B'C", 0 despite the fact that we chose A to be 0? If so, your claim is toast.

Get in the habit of ALWAYS asking if your results make sense. If you can't tell whether or not they make sense, then think about them that much harder because you will learn a lot in the process.

15. ### rtran2015 Thread Starter New Member

Oct 3, 2015
13
0
Oh okay so

ABC = 111
A'B'C' = 000

Due to the Axiom of A1 where [ B=0 if B != 1] and [B'=1 if B=0]

Hence
if ABC = 111, then A'B'C' = 000
= 1 + 0 = 1 through Axiom 5' "AND/OR"

16. ### WBahn Moderator

Mar 31, 2012
20,068
5,666
It is NOT sufficient to show that you get what you want for one particular choice of inputs. You have to PROVE that you get what you want for ALL possible combinations of inputs.

I, on the other hand, only need to demonstrate that you DON'T get what you want for at least ONE combination of inputs.

What is the result of ABC + A'B'C' if A=0 and B = 1 ?

What does that tell you about your claim that ABC + A'B'C' = 1?

17. ### rtran2015 Thread Starter New Member

Oct 3, 2015
13
0
oh alright,
Then it isn't sufficient to say ABC + A'B'C' is equal to 1
I will try a different approach to the question then

The given equation:

= ABCD' + A[BCD]' + ([A+B+C+D]')
= ABCD' + A(B'+C'+D') + ([A+B+C+D]'); T12 DeMorgan's
= ABCD' + A(B'+C'+D') + (A'B'C'D'); T12 DeMorgan's
= ABCD' + AB' + AC' + AD' + A'B'C'D'; T8 Distribute <---
-----------------------------------------------------------------------
=A(BCD'+B') + ...; T8
=A(B'+CD')+...; T6, T10
=AB'+ACD'+AC'+...;T8
-----------------------------------------------------------------------
=A[BCD]'+[A+B+C+D]'; T8,T12, T12

This is as far as I have gotten

18. ### WBahn Moderator

Mar 31, 2012
20,068
5,666
This is NOT the given equation!

The given equation is:

Which is

Y = ABCD' + AB'C'D' + (A + B + C + D)'

You are now trying to claim that

B'C'D' = (BCD)'

Again, ask yourself if that makes sense!

The left hand side is 0 if ANY of the three variables is 1. But on the right hand side, in order to be 0 ALL of the three variables must be 1.

Stop being so sloppy!

19. ### rtran2015 Thread Starter New Member

Oct 3, 2015
13
0
Bear with me on this

The given equation IS:
ABCD' + A[BCD]' + ([A+B+C+D]')

This equation is wrong:
Y = ABCD' + AB'C'D' + (A + B + C + D)'

I understand:
B'C'D' != (BCD)'
B'C'D' == (B+C+D)';T12 DeMorgan's

would it change if there was another variable:

A*(B'+C'+D') = A*(BCD)'; T12?

I apologize for any confusion
I'm still a beginner

= ABCD' + A[BCD]' + ([A+B+C+D]')
= ABCD' + A(B'+C'+D') + (A+B+C+D)'; T12 <---- Stop me if I am wrong here
= ABCD' + A(B'+C'+D') + A'B'C'D'; T12
= ABCD' + AB' + AC' + AD' + A'B'C'D'; T8
-----------------------------------------------------------------------
=AB' +ACD' + AC' +AD' +A'B'C'D';T8
=A(B'+C'+D')+A'B'C'D';T8
---------------------------------------------------
=A[BCD]'+A'B'C'D';T12
=A[BCD]'+[A+B+C+D]'; T12

20. ### WBahn Moderator

Mar 31, 2012
20,068
5,666
Okay, so if I understand you correctly you are saying that the original equation for part (c) in your original post was incorrect and that the actual given starting equation is:

ABCD' + A[BCD]' + ([A+B+C+D]')

Good!

Nope. You have A*(something) = A*(something) where you have merely expressed (something) using two different, but equivalent, forms.

Not a problem -- you learn this stuff by fighting with this stuff.

Fine to this point. Before proceeding as you do from here, see if you can't simplify this expression a bit more directly.

For instance, what does ABCD' + AD' simply to?

Without saying whether this is valid or not, there is the question of what it means to "simplify" an expression. There really needs to be some metric by which you can determine that one expression is "simpler" than another. In problems like this, it is usually implied that "simplest" means a SOP form with as few terms as possible.