# Boolean simplification....kind of stuck on a certain form of expression

#### RyanKim

Joined Sep 18, 2011
37
Hi guys I was wondering if anyone could take a look at 2 expressions I have that I can't seem to simplify. Im basically one step away from the answer at both junctures but I can't logically apply a law that I know how to use to get the answer.

The first equation that I've pretty much solved has boiled down to this:

1.) A!B! + B!C + A! + B! + C! (Where ! means not. E.g A! = A not)

Final answer suppose to be A! + B! + C!

and secondly

2.) Q! + R! + S! + R!S + QR!R!

Final Answer suppose to be Q! + R! + S!

I cant seem to figure out what to do next. Im not aware of any law that can be applied here...though I do notice in both expressions B! and R! may have something to do with it respectively?

Thanks!

#### t_n_k

Joined Mar 6, 2009
5,455
A!B! + B!C + A! + B! + C!

A!(B! + 1) + B!(C + 1) + C!

what's next?

#### t_n_k

Joined Mar 6, 2009
5,455
I don't think your second question partial solution is correct - if that's the given solution. Can you post the original question?

#### justtrying

Joined Mar 9, 2011
435
for the second part try factoring out R!

Q!+S!+R!(1+S+QR!) =

is there any law you can apply now? think of 1+X = X

p.s. even before grouping, you can simplify R!R! - recall XX = X

#### RyanKim

Joined Sep 18, 2011
37
Ahh thanks T_N_K I think I got the first bit...(X + 1) = 1. For the second part perhaps I should have added brackets but that was the given solution by my prof.

Q! + R! + S! + (R!S + QR!R!)

So I guess R!R! = R! therefore (R!S + QR!) = R! ? Then Q! + R! + S! + (R!) = Q! + R! + S! Since ( X + X = X)?

f = ( (QRS) (R + ( S + (QR!) )! )!

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#### t_n_k

Joined Mar 6, 2009
5,455
A computer generated minimization gives me the solution

X = not Q or not R or not S or not RS;

which doesn't match up with the solution you have.

#### RyanKim

Joined Sep 18, 2011
37
Really now? Hmm thats odd...Well if ud like to see my profs solutions for them Ill upload a PDF of it.

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#### justtrying

Joined Mar 9, 2011
435
you can also do this:

!(QRS(R+!(S+Q(!R))) = !(QRS(R+!S(!Q+R)) = !(QRS(R+!S!Q+!SR) = !(QRS)

if you are wondering about how I got the final answer, check out basic multiplication rules

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#### justtrying

Joined Mar 9, 2011
435
Ahh thanks T_N_K I think I got the first bit...(X + 1) = 1. For the second part perhaps I should have added brackets but that was the given solution by my prof.

Q! + R! + S! + (R!S + QR!R!)

So I guess R!R! = R! therefore (R!S + QR!) = R! ? Then Q! + R! + S! + (R!) = Q! + R! + S! Since ( X + X = X)?

f = ( (QRS) (R + ( S + (QR!) )! )!
not quite...

(R!S + QR!) cannot equal R! but Q!+R!+S!+R!S+QR! is valid. Then what happens if you factor out R! (see earlier post)

#### t_n_k

Joined Mar 6, 2009
5,455
Really now? Hmm thats odd...Well if ud like to see my profs solutions for them Ill upload a PDF of it.
My mistake. Finger trouble - I didn't enter the equation correctly. The application now spits out your given answer.