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EDIT: My apologies if I started this thread in the wrong forum!

Z = P.Q.R + P.R

Where

. = AND

+ = OR

~ = NOT

A+B = B+A, A.B = B.A

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

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

A+A = A, A.A = A

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

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

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

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

## 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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