Hello,
I'm trying to solve Boolean Algebra Problem . And here's it
Prove that X'Y'Z' + X'YZ + XY'Z + XYZ' = (X xor Y)' xor Z
I've proofed that
this problem = (X Xor Y xor Z)' but i can't get that one above . Can anybody help,please?
Hello,
I'm trying to solve Boolean Algebra Problem . And here's it
Prove that X'Y'Z' + X'YZ + XY'Z + XYZ' = (X xor Y)' xor Z
I've proofed that
this problem = (X Xor Y xor Z)' but i can't get that one above . Can anybody help,please?
One way to prove such a statement is to set it up a "truth table". Since each of the two elements, x, y, and z, can be either True or False, there are $\displaystyle 2^3= 8$ cases. If x= y= x= False, then x'= y'= z'= True so x'y'z is True while x'yz, x'yz, and xyz' are false. True+ False+ False+ False= True. The left side is True in this case. On the right, x xor y' is "False xor True= True" and so the right side becomes "True xor False= True".
Do the same with the other 7 cases and see if you get both sides the same in each case.
This is not the complete list of postulates dealing with xor. This is the definition of a connective xnor. If I replace all xnor's with the right-hand side, I'll be left with xor's and the usual connectives, but I have to way to relate xor to those connectives.
Also, in post #1 you said that you proved X'Y'Z' + X'YZ + XY'Z + XYZ' = (X xor Y xor Z)'. You could not have done this without some laws about xor. Obviously, if xor is an arbitrary connective, this equality is not true in general.
Yup i know that I know that A xor B = A'B + AB' and a xnor b = A'B' + AB
X(Y+Z) = XY + XZ and so on. i didn't expect that I've to mention that
I'm trying to figure out if what I've done is right how can i put (X xor Y xor Z)' = (X xor Y)' xor Z
It is true that (x xor y)' = x' xor y = x xor y'. Intuitively, x xor y says that x ≠ y, so (x xor y)' says that x = y. Similarly, x' xor y says that x' ≠ y, i.e., x = y. Formally, the laws above are easy to prove using X xor Y = XY' + X'Y, (X xor Y)' = XY + X'Y' and X'' = X.
Using these laws, (X xor Y xor Z)' = ((X xor Y) xor Z)' = (X xor Y)' xor Z.