# Boolean algebra simplification help

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• Jan 15th 2009, 04:19 AM
^_^Engineer_Adam^_^
Boolean algebra simplification help
Simplify the ff. Boolean function using De morgan's identity and or boolean identity and or special logic functions:
note ' is the complemented value

a.)F(w,x,y,z) = [(y+x)(w+z)']' ==> Ans: (w'x' + w + z)
= (y+x)' + (w+z) de morgan
= y'x' + w + z ????

b.)F(X,Y,Z) = [(X + Y' + Z')(X + Z')]' ==>Ans: X'YZ
= (X + Y' + Z')' + (X + Z')' de morgan
= X'(Y'+Z')' + (X'Z) de morgan
= X'YZ + X'Z
= X'Z(Y+1)
= X'Z ????

c.) F(x,y,z) = x'yz' + x'yz + (xyz' + xyz)' ==> Ans: y
= x'y(z' + z) + (xy(z' + z))'
= x'y + (xy)'
= x'y + x' + y'
= x' + y' ?????

are the answers obtainable through simplifying or is it just a typo?
• Jan 15th 2009, 05:01 AM
Moo
Quote:

Originally Posted by ^_^Engineer_Adam^_^
Simplify the ff. Boolean function using De morgan's identity and or boolean identity and or special logic functions:
note ' is the complemented value

a.)F(w,x,y,z) = [(y+x)(w+z)']' ==> Ans: (w'x' + w + z)
= (y+x)' + (w+z) de morgan
= y'x' + w + z ????

b.)F(X,Y,Z) = [(X + Y' + Z')(X + Z')]' ==>Ans: X'YZ
= (X + Y' + Z')' + (X + Z')' de morgan
= X'(Y'+Z')' + (X'Z) de morgan
= X'YZ + X'Z
= X'Z(Y+1)
= X'Z ????

c.) F(x,y,z) = x'yz' + x'yz + (xyz' + xyz)' ==> Ans: y
= x'y(z' + z) + (xy(z' + z))'
= x'y + (xy)'
= x'y + x' + y'
= x' + y' ?????

that's weird, I think you're correct for all of them (Surprised)
Or I'm mistaking all the way, because 3 typos may be too much...
• Jan 15th 2009, 05:12 AM
Isomorphism
Quote:

Originally Posted by ^_^Engineer_Adam^_^
Simplify the ff. Boolean function using De morgan's identity and or boolean identity and or special logic functions:
note ' is the complemented value

a.)F(w,x,y,z) = [(y+x)(w+z)']' ==> Ans: (w'x' + w + z)
= (y+x)' + (w+z) de morgan
= y'x' + w + z ????

The answer is obviously a typo. You can easily see that if you substitute w=z=x=0 and y=1. However it does not mean your answer is right.... But you have justified every step correctly.. so you are right (Clapping)

Quote:

Originally Posted by ^_^Engineer_Adam^_^
b.)F(X,Y,Z) = [(X + Y' + Z')(X + Z')]' ==>Ans: X'YZ
= (X + Y' + Z')' + (X + Z')' de morgan
= X'(Y'+Z')' + (X'Z) de morgan
= X'YZ + X'Z
= X'Z(Y+1)
= X'Z ????

Again a typo... Choose x=y=0 and z=1.

Quote:

Originally Posted by ^_^Engineer_Adam^_^
c.) F(x,y,z) = x'yz' + x'yz + (xyz' + xyz)' ==> Ans: y
= x'y(z' + z) + (xy(z' + z))'
= x'y + (xy)'
= x'y + x' + y'
= x' + y' ?????

are the answers obtainable through simplifying or is it just a typo?

Try y=0 and x=1. Clearly your original expression is 1. But their answer says 0.

As a side note, this type of expressions functional verification is automated these days. Programs like Verilog can automatically test with certain inputs. If you are wondering how i generated those combination of values.... I was just trying to give inputs that makes the expected answer differ from your answer. Which is the same as saying I was trying to make expected answer as the complement of your answer or in other words, expected answer XOR your answer = 1.

For example in the second problem:

(X'YZ) XOR (X'Z) = 1 is the same as X'Z (Y XOR 1) = 1 is the same as X'Y'Z = 1. Thus I chose X=Y=0 and Z=1 ;)
• Jan 15th 2009, 06:16 AM
^_^Engineer_Adam^_^
these are the 3 out of 20 problems im getting confused at im sure it were typos

thanks alot for the help