# How Can I Simplify This Boolean Expression Using XOR and XNOR Functions?

• TheTopGun
In summary, the goal is to simplify a boolean expression using XOR and XNOR functions, but the last two terms get in the way. They are trying to simplify it further using a 4 variable truth table but have not yet succeeded.
TheTopGun

## Homework Statement

Hey there, I'm having trouble simplifying a boolean expression using XOR and XNOR functions.
The final goal is to draw a logic circuit for the expression using NAND and XOR gates only.

## Homework Equations

Assuming
W' = Not WW' X' Y' Z' + W' X' Y Z + W' X Y' Z + W' X Y Z' + W X Y' Z' + W X' Y' Z

## The Attempt at a Solution

W' X' (Y⊕Z)' + W' X (Y⊕Z) + ...

So far if I'm working correctly (?) I can simplify the first four expressions, It's just the last two that get me as they have no like terms ?

Once I understand this I should have no troubles drawing the logic circuit.

Thankyou for any responses

you could try drawing a truth table for the terms and see if you spot a pattern that matches XOR or NAND

like X xor Y'

Hmm okay thanks,

I've come up with this

W' X' (Y⊕Z)' + W' X (Y⊕Z) + W Y' (X⊕Z)

Is this on the right track..before I simplify further.

W x y' z' + w x' y' z = w y'❲ ... ❳

wow, PF has such a nanny editor that it converted my upper case to lower!

TheTopGun said:
Hmm okay thanks,

I've come up with this

W' X' (Y⊕Z)' + W' X (Y⊕Z) + W Y' (X⊕Z)

Is this on the right track..before I simplify further.

That seems right. Once complete you could do a truth table on it and see if it agrees with original.

TheTopGun said:
Hmm okay thanks,

I've come up with this

W' X' (Y⊕Z)' + W' X (Y⊕Z) + W Y' (X⊕Z)

Is this on the right track..before I simplify further.
You can see how to simplify this further? If not, I suggest that you construct a 4 variable Truth Table. Also construct the Truth Table for your simplified expression, and with any luck there may be some correspondence.

I haven't tried it.

For future reference (im not sure why anyone would care lol, but anyway), the equation can be simplified down to:

W' (X⊕Y⊕Z)' + Y' (W⊕X⊕Z)' ...where (A⊕B⊕C)' represents a XNOR gate.

This was obtained using karnough maps but I am sure there are easier ways...

## 1. What is Boolean Algebra Simplification?

Boolean Algebra Simplification is a method used to reduce and simplify complex Boolean expressions to their most simplified form. It is based on the principles of Boolean logic, which deals with the manipulation of binary variables and logical operations such as AND, OR, and NOT.

## 2. Why is Boolean Algebra Simplification important?

Boolean Algebra Simplification is important because it allows us to simplify and analyze complex logic circuits and expressions, making them easier to understand and work with. It is also essential in the design and optimization of digital systems, such as computers and electronic devices.

## 3. What are the basic rules of Boolean Algebra Simplification?

The basic rules of Boolean Algebra Simplification include the commutative, associative, distributive, and absorption laws. The commutative law states that the order of the operands does not affect the result of the logical operation. The associative law states that the grouping of operands does not affect the result. The distributive law states that AND and OR operations can be distributed over each other. The absorption law states that an expression and its simplified form are equivalent.

## 4. How do you simplify a Boolean expression using Boolean Algebra?

To simplify a Boolean expression using Boolean Algebra, you can follow these steps:

1. Use the basic rules of Boolean Algebra to simplify the expression.
2. Use De Morgan's laws to eliminate any NOT operations.
3. Combine any like terms or variables.
4. Eliminate any unnecessary parentheses.
Once these steps are completed, you will have a simplified Boolean expression.

## 5. What are some common mistakes to avoid when simplifying Boolean expressions?

Some common mistakes to avoid when simplifying Boolean expressions include forgetting to apply the basic rules of Boolean Algebra, incorrectly applying De Morgan's laws, and not distributing operations properly. It is also important to be careful when combining like terms and double-check the final simplified expression to ensure it is equivalent to the original expression.

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