Jump to content
Sign in to follow this  
xxHunterRosexx

Boolean algebra proof two expressions are equivalent

Recommended Posts

Hi,

I’m not sure if I’m posting in the right forum but please let me know if I’m not.

I’m practicing for the test and trying to prove that two boolean algebra expressions are equivalent:

x1=abc+bc+ac+abcx1=a′b′c+bc′+ac+ab′c

x2=bc+bc+abx2=b′c+bc′+ab

I got up to here:

LHS

abc+bc+ac+abca′b′c+bc′+ac+ab′c

RHS

=(a+a)c+bc+ab=(a+a′)c+bc′+ab

=abc+abc+bc+ab=ab′c+a′b′c+bc′+ab

=abc+bc+abc+ab(c+c)=a′b′c+bc′+ab′c+ab(c+c′)

=abc+bc+abc+abc+abc=a′b′c+bc′+ab′c+abc+abc′

=abc+bc+abc+(b+b)abc+abc=a′b′c+bc′+ab′c+(b+b′)abc+abc′

=abc+bc+abc+abbc+abbc+abc=a′b′c+bc′+ab′c+abbc+ab′bc+abc′

The next step is supposed to be

=abc+bc+abc+ac+0+abc

I do not see how they got to that step. If someone give a pointer to what I should be doing next for this theorem, I would be beyond grateful.

Share this post


Link to post
Share on other sites

Well the zero comes from b ANDed with NOT b.

Not seeing how they cancel out the bb in abbc. Would think one should be left.

Share this post


Link to post
Share on other sites

You may have a typo. 

x1=a′b′c+bc′+ac+ab′c=b'c+bc'+ac

x2=b′c+bc′+ab

They are the same, except for the third term in each.

Share this post


Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
Sign in to follow this  

×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.