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!

 

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+....

Then I will help you.

Ratch

 

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

 

Again any help appreciated the more i look at the problem the harder it is to figure out.

 

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'

 

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.

 

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.

http://forum.allaboutcircuits.com/showthread.php?t=22347&highlight=logic+lovers

Ratch

 

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.

Your second posted answer I agree with: p'qr's'+pqr's+pq's'+prs'

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

 

I'm such a tool. I totally misread my own K-map. Shouldn't be doing boolean algebra after midnight!

 

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