# Thread: Boolean Algerba Problem

1. ## Boolean 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?

2. ## Re: 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.

3. ## Re: Boolean Algerba Problem

Originally Posted by HallsofIvy
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.

I want to solve that problem using postulates

Such that

Z'(X xor y) ' + Z(X xor Y) Let X Xor y = A

Z'A' + ZA = (Z xor A)' = (Z xor A xor Y)'

4. ## Re: 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.

5. ## Re: Boolean Algerba Problem

Originally Posted by emakarov
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.
X xnor Y = (X xor Y)'

6. ## Re: 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.

7. ## Re: Boolean Algerba Problem

Originally Posted by emakarov
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.

That's all i know postulates about xor and xnor

8. ## Re: 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.

9. ## Re: 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.

10. ## Re: 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

11. ## Re: 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.

12. ## Re: Boolean Algerba Problem

Originally Posted by emakarov
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.
Wow Thanks alot!!