# Boolean Algebra Simplification

1. Apr 30, 2013

### TheTopGun

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

Assuming
W' = Not W

W' X' Y' Z' + W' X' Y Z + W' X Y' Z + W' X Y Z' + W X Y' Z' + W X' Y' Z

3. 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

2. Apr 30, 2013

### Staff: Mentor

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'

3. Apr 30, 2013

### TheTopGun

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.

4. Apr 30, 2013

### Staff: Mentor

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!

5. Apr 30, 2013

### Staff: Mentor

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

6. Apr 30, 2013

### Staff: Mentor

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.

7. May 1, 2013

### TheTopGun

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 Im sure there are easier ways...