Simplify boolean expression (a+b+c)*(a'+c)*(a'+b')

[dba]

Homework Statement

I need to simplify the boolean expression by algebraic manipulation as much as possible.
(a+b+c)*(a'+c)*(a'+b')

Homework Equations

+ stands for OR
* stands for AND (or no operator between variables)
' Stands for NOT

The Attempt at a Solution

I tried but I do not know if this is correct.

(a+b+c)*(a'+c)*(a'+b')
= [aa' + ac + ba' + bc + ca' + cc] * (a'+b') => aa'=0, cc=c
= [0 + ac + ba' + bc + ca' + c] * (a'+b') => c + bc = c
= [ac + ba' + ca' + c] * (a'+b') => c + ac = c
= [ba' + ca' + c] * (a'+b') => c + a'c = c not sure if that works
= [ba' + c] * (a'+b')
= ca' + cb' + a'a' + a'b' + ba' + bb' => a'a'=a', bb'=0
= ca' + cb' + a' + a'b' + ba' => factor out the last two a'
= ca' + cb' + a' + a'(b' + b) => b'+b=1
= ca' + cb' + a' + a' => a'+a'=a'
= ca' + cb' + a' => factor out a'
= a'(1+c) + cb' => 1+c = 1
= a+cb'

Thanks for any help!

 

[NascentOxygen]

(a+b+c)*(a'+c)*(a'+b')
= [aa' + ac + ba' + bc + ca' + cc] * (a'+b') => aa'=0, cc=c
= [0 + ac + ba' + bc + ca' + c] * (a'+b') => c + bc = c
= [ac + ba' + ca' + c] * (a'+b') => c + ac = c
= [ba' + ca' + c] * (a'+b') => c + a'c = c not sure if that works
= [ba' + c] * (a'+b')

I think you're right this far, but I can't follow your working thereafter.

To check your own work, you can draw up a truth table. The expression you are given should always evaluate to the same values as your "simplified" version. If in an exam, you don't have time for a complete TT, try just a few values, e.g., a=b=1 c=0
then original expression is 0; your expression evaluates to 1.