I'm attempting to simplify this and don't know if I'm doing it right.
This is the problem: (a+b+c')(a'b'+c)
Attempted solution:
((a+b)+c') ((a+b)'+c) By duality
(a+b)(a+b)' + (a+b)c + (a+b)'c' + cc' Distributive
0 + (a+b)c + (a+b)'c'+ 0
Final answer: 1 . Is it right to say (a+b)c + (a+b)'c' is equal to 1? I just thought of it as being no different than x + x' = 1 but I have a suspicion this isn't the right step here. Any help appreciated.
1 Answer
First of all, no $(a+b)c+(a+b)'c'\not\equiv 1$. If a and b are 1 but c is 0, this expression evaluates to 0. This is actually the xnor of $(a+b)$ and $c$.
Simplifying boolean expressions isn't so different to numeric algebra. Expand brackets and then group like terms. Although we do add a step for removing contradictory terms.
You should try expanding AND'ed brackets first:
$$(a+b+c')(a'b'+c)=a(a'b'+c)+b(a'b'+c)+c'(a'b'+c)$$ $$=aa'b'+ac+ba'b'+bc+c'a'b'+c'c$$
And then you remove terms which contain contradictions($aa'$ is 0): $$aa'b'+ac+ba'b'bc+c'a'b'+c'c=ac+bc+c'a'b'$$
Then you would check for redundant terms. This one doesn't have any but for argument's sake, lets say you had $ac+abc$. Since $ac\to abc$, the $abc$ would be unnecessary. Therefore $ac+abc=ac$. This can sometimes be more of an art than a science, for example $ac+a'bc=ac+bc$ and $ab+a'b=b$.
Otherwise, you can always fall back on Karnaugh maps if you feel like it might still not be simplified enough.
