A smal problem with boolean algebra

[gipc]

There's something I can't seem to figure out.

say we have the function
f=xz +x'z'

why is this equals (x xor z)' ?
i thought using De-Morgan's law we shall get

(x xor z)'=(x'z+z'x)'=(xz'zx')=0

then why does (x xor z)'=xz+x'z' ?


and one more small thing,
say we have
(x xor y) xor (x xor z)'
why does (x xor y) xor (x xor z)' = (x xor y xor x xor z)'

how did they come to this term?

 

[vela]

De Morgan's law also tells you (ab)' = a' + b', so

(x xor z)' = (x'z+z'x)' = (x'z)'(z'x)' = ...

I'll let you finish it.

 

[gipc]

thanks so much i feel so stupid for missing the obvious.

one more thing though,
why does the following equation stand?

(x XOR y) XOR (x XOR z)' = (x XOR y XOR x XOR z)'

 

[vela]

That's kind of like asking why a trig identity is true. It's true because it's a consequence of how things are defined. You just have to prove it by using algebraic manipulations or just listing all possible values of x, y, and z and showing the two sides give the same results.

 

[DeeNos]

It seems like you are having trouble understanding the concept of xor (exclusive OR) in boolean algebra. Let's break it down step by step:

1. First, let's look at the function f = xz + x'z'. This function can also be written as f = xz + x'z' + 0. This is because 0 is the identity element for the boolean addition operation, which means that adding 0 to any boolean expression does not change its value. So, we can rewrite f as f = xz + x'z' + 0.

2. Now, let's use De-Morgan's law to simplify this expression. De-Morgan's law states that (A + B)' = A'B'. Applying this law to our expression, we get f = (xz)'(x'z')' + 0. Using De-Morgan's law again, we get f = (x' + z')(x + z) + 0. This simplifies to f = x'z' + xz + xz' + z'z + 0.

3. Notice that xz + xz' + z'z is equal to 0, because z and z' are complements of each other, meaning that they both can't be 1 at the same time. So, we can rewrite f as f = x'z' + xz + 0. This is where xor comes into play. Xor is the boolean operation that means "either A or B, but not both." So, in this case, we can rewrite f as f = x xor z.

4. Now, let's look at the expression (x xor y) xor (x xor z)'. This can also be written as (x xor y) xor (x xor z) + 0. Applying De-Morgan's law, we get (x' + y')(x' + z') + 0. Simplifying this, we get (x'x' + x'z' + x'y' + y'z') + 0. Again, notice that x'x' + x'y' + y'z' is equal to 0. So, we are left with (x'z' + x'z') + 0. This simplifies to (x' xor z') xor 0, which is equal to x' xor z'.