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?

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- Jun 2nd 2012, 07:10 AMahmedzoro10Boolean Algerba Problem
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? - Jun 2nd 2012, 07:23 AMHallsofIvyRe: Boolean Algerba Problem
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. - Jun 2nd 2012, 08:35 AMahmedzoro10Re: Boolean Algerba Problem
- Jun 2nd 2012, 11:01 AMemakarovRe: Boolean Algerba Problem
Could you give the list of available postulates, especially concerning xor? This connective is less common than and, or, and not, so the postulates for it are less standard.

- Jun 2nd 2012, 01:40 PMahmedzoro10Re: Boolean Algerba Problem
- Jun 2nd 2012, 01:45 PMemakarovRe: Boolean Algerba Problem
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.

- Jun 2nd 2012, 01:47 PMahmedzoro10Re: Boolean Algerba Problem
- Jun 2nd 2012, 01:49 PMemakarovRe: Boolean Algerba Problem
How do your sources introduce xor? If you want to prove the original equality symbolically, I would expect that you have at least something like X xor Y = XY' + X'Y.

- Jun 2nd 2012, 01:53 PMemakarovRe: Boolean Algerba Problem
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.

- Jun 2nd 2012, 02:10 PMahmedzoro10Re: Boolean Algerba Problem
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 - Jun 2nd 2012, 02:22 PMemakarovRe: Boolean Algerba Problem
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. - Jun 2nd 2012, 02:52 PMahmedzoro10Re: Boolean Algerba Problem