Boolean algebra proof two expressions are equivalent

[xxHunterRosexx]

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=a′b′c+bc′+ac+ab′cx1=a′b′c+bc′+ac+ab′c

x2=b′c+bc′+abx2=b′c+bc′+ab

I got up to here:

LHS

a′b′c+bc′+ac+ab′ca′b′c+bc′+ac+ab′c

RHS

=(a+a′)c+bc′+ab=(a+a′)c+bc′+ab

=ab′c+a′b′c+bc′+ab=ab′c+a′b′c+bc′+ab

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

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

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

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

The next step is supposed to be

=a′b′c+bc′+ab′c+ac+0+ab′c

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.

[Endy0816]

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.

[mathematic]

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.