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' ?????