# Basic manipulating boolean identities

Discussion in 'Homework Help' started by Tjvelcro, Sep 21, 2012.

1. ### Tjvelcro Thread Starter New Member

Sep 21, 2012
2
0
Hi guys,

I have a very basic question regarding bollean identites and manipulating them.

Given a 2 input XOR gate, change the SOP expression to the POS expression using the identites.

So I have...

SOP F = A'B +AB'
POS F = (A+B)(A'+B')

I'm given the formulas similar to the http://sandbox.mc.edu/~bennet/cs110/boolalg/rules.html website.

I cannot seem to find a formula that completely matches what I have. I tried multiplying out the POS to get F = AB' + A'B but not sure if that is correct and not sure how I have go from SOP to POS. ANy help would be appeciated.

Tjvelcro

2. ### WBahn Moderator

Mar 31, 2012
18,087
4,917
Not surprising. The rules are the tools. It's like giving a scupltor a chunk of marble and stoneworking tools and having them say that none of them completely match the bust they are supposed to create. You will generally use several rules in a carefully selected order in order to incrementally move from your starting point to your desired goal. Just like the sculptor.

Not sure if what is correct? Multiplying it out to get F=AB'+A'B? Well, if you multiply it out and get it into the form you need, then it is correct.

If you figure out how to go one way, then just reverse the steps to go the other. Often times this is the easiest way to solve a problem -- start with the desired solution and work backward to get to what you are starting with. It's like working one of those maze puzzles you see on restaurant placemats.

For any more detailed assistance, you need to present what you have done up to this point so that we have a starting point from which to guide you.

3. ### Tjvelcro Thread Starter New Member

Sep 21, 2012
2
0
Right, I did try to start at the end so to speak. I started with the POS form of XOR and multipled it out.

POS F = POS F = (A+B)(A'+B') = 0 + AB' + A'B + 0

It looks just like the SOP F = A'B +AB' which I started with. It would be great, however I need to specify how to go from SOP to POS and what identities to use.

The only thing I could think of was....
F = A'B +AB' = AA' + AB' A'B + + BB' = A(A'+B')+B(A'+B')... can is just change to this...? (A+B)(A'+B')? If so, I'm not sure what identity allows it to do that. Perhaps its just a normal collect the alike terms or something.

Tjvelcro

4. ### WBahn Moderator

Mar 31, 2012
18,087
4,917
You've got it!

F = A'B + AB'
F = 0 + A'B + AB' + 0 (Identity)
F = A'A + A'B + AB' + BB' (Indempotence)
F = A'(A+B) + (A+B)B' (Distributivity)
F = (A+B)A' + (A+B)B' (Commutivity)
F = (A+B)(A'+B') (Distributivity)

Remember, distributivity applies to any factor, not matter how complex. So

(Anything)(AnythingElse + YetSomethingElse)

is equal to

(Anything)(AnythingElse) + (Anything)(YetSomethingElse)