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