Simplification of Boolean Expression

Thread Starter

Sky28

Joined Jul 29, 2010
3
Hi there,
Que: Simply the boolean expressions to a minimum number of literals:
( x + y + z') ( x'+y'+z)
Solution:
( xy'+x'y) + ( xz+x'z') + ( yz+y'z')
= x( XOR symbol) y + x( XNOR symbol) + y (XNOR symbol) z

i believe this is the solution but would like to double confirm. Affirmative that this is the solution?
For those who took the liberty to check thank you so much.
 

mps

Joined Feb 4, 2010
8
Hi there,
Que: Simply the boolean expressions to a minimum number of literals:
( x + y + z') ( x'+y'+z)
Solution:
( xy'+x'y) + ( xz+x'z') + ( yz+y'z')
= x( XOR symbol) y + x( XNOR symbol) + y (XNOR symbol) z

i believe this is the solution but would like to double confirm. Affirmative that this is the solution?
For those who took the liberty to check thank you so much.

1)
( x + y + z') ( x'+y'+z)
solving it we get
xy'+xz+yx'+yz+x'z'+y'z
z+x'z'+yx'+xz+xy'
z(1+x)+x'z'+x'y+xy'
(x+x')z+x'(z'+y)+xy'
x'+xz+x'y+xy'
x'+xz+xy'

Lol
the solution seems to be fishy.
I have reduced it to lesser than the given answer
 
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