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Simplifying Boolean Expressions using the Laws of Boolean Algebra

  • Thread starter revelator
  • Start date
  • #1
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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
 
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Answers and Replies

  • #2
33,173
4,858
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
22
0
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
33,173
4,858
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.
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
22
0
??? 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
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
33,173
4,858
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
22
0
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
33,173
4,858
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
22
0
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
33,173
4,858
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
22
0
Right, so Z = PRQ + PR = PR(Q + 1) = PR(1) = 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
Ahh ok, thank you so much for clearing that up! It`s been driving me batty.

One more quick question, if y`all dont 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
33,173
4,858
It might be reasonable to apply an identify first. I think you have to take things on a case-by-case basis.
 
  • #14
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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
33,173
4,858
Sure. If there are obvious simplifications, make them.
 
  • #16
22
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Sure. If there are obvious simplifications, make them.
Awesome, thank you again so much, you were a great help to me :)
 

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