# Boolean Algebra Simplification Help Needed

#### wordy007

Joined Dec 30, 2009
6
Hi im new here and have never used forums like this but so far it looks very good. Im having a spot of trouble simplifying a boolean algebra expression for my homework and tried everything i can think of but nothing seems to look right.

The expression is: (p' = inverted) (++ = XOR)
(p'qr's') + (pq'r's') + (pq'rs') + (pqr's)+(pqrs')

Things i have tried so far:
Distributive rule:
(p'qr's') + pq's'(r'r) + pq(r'd+cd')
Cancellation ([r]*r) = 1:
(p'qr's') + pq's' + pq(r'd+cd')
XOR - (r'd+cd'):
(p'qr's') + pq's' + pq(c++d)

Then that is as far as i can get it.

I have also tried multiplying out the brackets at the beginning and using commutative law and cancellation to get to:
q'+sr'+ps'+q'rp+s'q+p+r
I would write the steps i took to get to this but its 3 pages long and more than likley i've done it all wrong.

Would appreciate any help i can get as i just can't get it into it's simplest form!

Last edited:

#### Ratch

Joined Mar 20, 2007
1,070
wordy007,

No one like to read that clunky, clumsy notation you are using. Who taught you that, and where does it come from? I see it often on questions about Boolean algebra. Resubmit your expression using the following notation. It is so much easier to read.

[p]*q*[r]*) ===> p'q'r's+pqrs+....

Ratch

#### wordy007

Joined Dec 30, 2009
6
Sorry about that was just copying what i saw others were doing, i'll edit it now and again any help at all appreciated

#### wordy007

Joined Dec 30, 2009
6
Again any help appreciated the more i look at the problem the harder it is to figure out.

#### Fraser_Integration

Joined Nov 28, 2009
142
I've just put it into a Karnaugh map (information in the online textbook if you need it), and it doesn't simplify much, I came out with:

P'QRS + PQRS' + PQ'S' + PRS'

#### JotDot

Joined Jan 1, 2010
3
I Haven't seen the notation specified by Ratch - I prefer what you submitted.

Sorry Fraser_Integration but I can't seem to agree with your results - not saying I messed up

Original Expression:
(p'qr's') + (pq'r's') + (pq'rs') + (pqr's)+(pqrs')

factor out the r's' and the rs' we get:
(r's') (p'q + pq') + (rs')(pq' + pq) + pqr's

Rewriting (pq' + pq) will give (p)(q'+q) or simply p
Leaving us with (r's') (p'q + pq') + prs' + pqr's

Note that p'q + pq' is (p xor q) but that does not give you much.

Thus the final form would be a sum of four products:
p'qr's' + pq'r's' + pqr's + pr's

Making a Karnaugh Map for this shows there is not much possible.

Last edited:

#### Ratch

Joined Mar 20, 2007
1,070
Fraser_Integration,

I've just put it into a Karnaugh map (information in the online textbook if you need it), and it doesn't simplify much, I came out with:

P'QRS + PQRS' + PQ'S' + PRS'
Wrong answer if you are using JotDot's intrepretation of the original Boolean expression. See below.

JotDot,

I Haven't seen the notation specified by Ratch - I prefer what you submitted.
I don't understand. First you say you say you have not seen the notation I recommended, and then you use it even though you prefer the OP's notation. Anyway, the answer you submitted is wrong. The correct answer is p'qr's'+pqr's+pq's'+prs' . How do I know that is the correct answer? I used a K-map and checked it with the program featured in the link below.

Ratch

#### JotDot

Joined Jan 1, 2010
3
Sorry. I didn't realize the O/P edited the orignal post - I was expecting a second post so I wrongly assumed you didn't like how it was written. I should pay better attention.

I was simply trying to work out the math for the O/P to show the simplifications - which is what I believe the O/P wanted - not just the final answer (which I disagreed with).
I see I goofed writng down the last poduct; misplacing a symbol. It should be +prs' (not +pr's) Also I did miss one final reduction when I now look at the K-Map again.

p.s.: That progam you linked is great for double checking results Would have been handy for me before I posted

#### Fraser_Integration

Joined Nov 28, 2009
142
I'm such a tool. I totally misread my own K-map. Shouldn't be doing boolean algebra after midnight!

#### wordy007

Joined Dec 30, 2009
6
thanks very much guys for the help, it seems this boolean algebra doesn't simplify down much. Thank you all very much again all i have to do now is turn that into a logic circuit