Simplifying Boolean Expressions using the Laws of Boolean Algebra

In summary: I think I get it now, the identity I can use is A+AB=A, which would make the final answer Z = PR? That's right. The rest of it was just fluff.
  • #1
revelator
25
0
EDIT: My apologies if I started this thread in the wrong forum!

Homework Statement


Z = P.Q.R + P.R

Where
. = AND
+ = OR
~ = NOT

Homework Equations



(Commutative Law)
A+B = B+A, A.B = B.A

(Associate Law)
(A+B)+C = A+(B+C), (A.B).C = A.(B.C)

(Distributive Law)
A(B+C) = AB+AC, A+(B.C) = (A+B).(A.C)

(Identity Laws)
A+A = A, A.A = A
A.B+A~B = A, (A+B).(A+~B) = A

(Redundancy Laws)
A+A.B = A, A.(A+B) = A
0+A = A, 0.A = 0
1+A = 1, 1.A = A
~A+A = 1, ~A.A = 0

(Demorgan's Theorem)
~(A+B) = ~A.~B, ~(A.B) = ~A + ~B

The Attempt at a Solution



I don't even know how to go about finding the solution. I'm not recognizing how I should apply the above laws of Boolean Algebra to simplify my problem statement. I'm thinking I must have missed something fundamental while in class. Any tips on how to proceed would be greatly appreciated!

Just to clarify, I'm not looking to be given the answer, just some tips on how to use the Laws to get to the solution. Thanks in advance! :D
 
Last edited:
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  • #2
revelator said:
EDIT: My apologies if I started this thread in the wrong forum!

Homework Statement


Z = P.Q.R + P.R

Where
. = AND
+ = OR
~ = NOT


Homework Equations



(Commutative Law)
A+B = B+A, A.B = B.A

(Associate Law)
(A+B)+C = A+(B+C), (A.B).C = A.(B.C)

(Distributive Law)
A(B+C) = AB+AC, A+(B.C) = (A+B).(A.C)

(Identity Laws)
A+A = A, A.A = A
A.B+A~B = A, (A+B).(A+~B) = A

(Redundancy Laws)
A+A.B = A, A.(A+B) = A
0+A = A, 0.A = 0
1+A = 1, 1.A = A
~A+A = 1, ~A.A = 0

(Demorgan's Theorem)
~(A+B) = ~A.~B, ~(A.B) = ~A + ~B

The Attempt at a Solution



I don't even know how to go about finding the solution. I'm not recognizing how I should apply the above laws of Boolean Algebra to simplify my problem statement. I'm thinking I must have missed something fundamental while in class. Any tips on how to proceed would be greatly appreciated!

Just to clarify, I'm not looking to be given the answer, just some tips on how to use the Laws to get to the solution. Thanks in advance! :D

I take it that you want to simplify Z = PQR + PR

The two expressions on the right have a common factor. Rewrite the right side in factored form. What do you get?
 
  • #3
Mark44 said:
I take it that you want to simplify Z = PQR + PR

The two expressions on the right have a common factor. Rewrite the right side in factored form. What do you get?

Ahh yes, I am looking to simplify Z = PQR + PR, sorry for not being clear on that!

After factoring the right side, this is what I`ve come up with.
Z = Q + PP + PR + RP + RR

Provided my factoring is correct, I believe I can use the Identity Law AA=A to get -
Z = Q + P + PR + RP + R

And then, using the Commutative Law AB=BA to get -
Z = Q + P + PR + PR + R

And again using the Identity Law A+A=A to get -
Z = Q + P + PR + R

Then using the Redundancy Law A+AB=A to get -
Z = Q + P + R
 
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  • #4
revelator said:
Ahh yes, I am looking to simplify Z = PQR + PR, sorry for not being clear on that!

After factoring the right side, this is what I`ve come up with.
Z = Q + PP + PR + RP + RR
? That's not factored. I have no idea how you got this.
revelator said:
Provided my factoring is correct, I believe I can use the Identity Law AA=A to get -
Z = Q + P + PR + RP + R

And then, using the Commutative Law AB=BA to get -
Z = Q + P + PR + PR + R

And again using the Identity Law A+A=A to get -
Z = Q + P + PR + R
 
  • #5
Mark44 said:
? That's not factored. I have no idea how you got this.

Aww crud, I`m sorry. I took the common terms PR, and tried applying the FOIL method (First, Outer, Inner, Last) to it. I thought that`s what factoring was D:.

I suppose what I need to understand, is what needs to be done, to factor Z=PQR+PR.
 
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  • #6
revelator said:
Aww crud, I`m sorry. I took the common terms PR, and tried applying the FOIL method (First, Outer, Inner, Last) to it. I thought that`s what factoring was D:.

I suppose what I need to understand, is what needs to be done, to factor Z=PQR+PR.

Try Z = PR (1 + Q)
 
  • #7
As written, PQR + PR is a sum of two terms. Rewrite this expression as a product of two factors.

For example 2xy + 2y = 2y(x + 1).
 
  • #8
Mark44 said:
As written, PQR + PR is a sum of two terms. Rewrite this expression as a product of two factors.

For example 2xy + 2y = 2y(x + 1).

Ok, I think I understand now. What I`m doing is, taking the common expression (in the above example it would be 2y) and dividing it out. So I get

2xy / 2y = x
2y / 2y = 1

Which leaves me with 2y(x+1).

Similarly with the problem I originally posted PQR+PR, I divide PR out of both sides (as it`s the common expression).
PQR / PR = Q
PR / PR = 1
Z=PR(Q+1)
 
  • #9
You're really going the long way around to go from Z = PRQ + PR to Z = PR(Q + 1).

All you need to say is Z = PRQ + PR = PR(Q + 1).

You should be able to do the dividing in your head - you don't need to write it down.

Now, what further simplification can you do? There is one identity in your first post in this thread that will be useful.
 
  • #10
Mark44 said:
You're really going the long way around to go from Z = PRQ + PR to Z = PR(Q + 1).

All you need to say is Z = PRQ + PR = PR(Q + 1).

You should be able to do the dividing in your head - you don't need to write it down.

Now, what further simplification can you do? There is one identity in your first post in this thread that will be useful.

Oh yea, I wouldn`t take the long way like that for my assignment, was just listing out the steps i was taking, so that if I made any mistakes it could be pointed out.

I`m thinking the relevant identity in this case would be 1+A=1. Using that I can turn
Z=PR(Q+1)
into
Z=PR(1)
then I can use the inverse of that, 1A=A to get
Z=PR

I hope I`ve done it correctly this time, haha.
 
  • #11
Right, so Z = PRQ + PR = PR(Q + 1) = PR(1) = PR

revelator said:
then I can use the inverse of that, 1A=A to get Z = PR
That's not the inverse. 1 is the multiplicative identity. I don't think there's a multiplicative inverse, but the additive inverse of A is ~A, since A + ~A = 1
 
  • #12
Mark44 said:
Right, so Z = PRQ + PR = PR(Q + 1) = PR(1) = PRThat's not the inverse. 1 is the multiplicative identity. I don't think there's a multiplicative inverse, but the additive inverse of A is ~A, since A + ~A = 1

Ahh ok, thank you so much for clearing that up! It`s been driving me batty.

One more quick question, if y`all don't mind.

Does factoring always come before simplifications, or are there cases in which the best course of action would be to apply an identify first, and do factoring after? I`m just curious, as in another problem, it seems that it would be best to do just that.
 
  • #13
It might be reasonable to apply an identify first. I think you have to take things on a case-by-case basis.
 
  • #14
Mark44 said:
It might be reasonable to apply an identify first. I think you have to take things on a case-by-case basis.

I`m thinking it would be applicable in the case of Z=(P+Q).(R+S)+P.(R+~R)
 
  • #15
Sure. If there are obvious simplifications, make them.
 
  • #16
Mark44 said:
Sure. If there are obvious simplifications, make them.

Awesome, thank you again so much, you were a great help to me :)
 

1. What is Boolean algebra?

Boolean algebra is a type of algebra that deals with logical statements and operations on them, specifically using the values of "true" and "false". It is used to simplify logical expressions and is commonly used in computer science and digital electronics.

2. What are the basic laws of Boolean algebra?

The basic laws of Boolean algebra are the commutative law, the associative law, the distributive law, the identity laws, the complement laws, and the idempotent laws. These laws help in simplifying complex Boolean expressions.

3. How do I simplify a Boolean expression using Boolean algebra?

To simplify a Boolean expression, you can use the basic laws of Boolean algebra to manipulate the expression. Start by simplifying the individual terms using the identity and complement laws, then use the associative, commutative, and distributive laws to group similar terms together and reduce the expression further.

4. Can Boolean algebra be applied to real-world problems?

Yes, Boolean algebra can be used to solve real-world problems, especially in the fields of computer science and digital electronics. It is used to design and analyze logical circuits, develop algorithms, and simplify complex decision-making processes.

5. Are there any tools or software to simplify Boolean expressions using Boolean algebra?

Yes, there are many online tools and software available that can help you simplify Boolean expressions using Boolean algebra. Some examples include Wolfram Alpha, Boolean Algebra Calculator, and Logic Friday. These tools can save time and effort in solving complex Boolean expressions.

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