Simplification of logic circuit using algebra

Question

The following is the logic circuit:

I have to simplify the following:

(((AB)')'+(B+C)+(AB)'(B+C)')C

=(AB+B+C+(A'+B')(B'C'))C

=(B+C+A'B'C'+B'C')C

=BC+C+A'B'C+B'C

=C+A'BC'+B'C

Answer 1

The expression you got from the circuit is not correct. It should be:

$(((AB)')'(B +C)+ (AB)'(B+C)')C$

By Double Negation and DeMorgan that gives you :

$(AB(B+C)+ (A'+B')B'C')C$

The $B$ absorbs the $B+C$, while the $B'$ absorbs to $A'+B'$:

$(AB+B'C')C$

Distribution:

$ABC+B'C'C$

And since the last term is $0$, you are left with:

$ABC$