# Simplification of logic circuit using algebra

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

• One quick way to check, to yourself, whether it's correct is to do a truth table. – Arthur Apr 9 at 18:55
• How did you reach the second line? Where did $ABC$ come into the picture? – Shubham Johri Apr 9 at 19:14
• Assuming the last line is correct, then you can simplify this to AC+B'C. But your second line does not look correct .. unless you miswrote the very starting expression. – Bram28 Apr 9 at 19:19
• I made some changes – Jarvis Ferns Apr 9 at 19:20
• Cross-posted: math.stackexchange.com/q/3181419/14578, electronics.stackexchange.com/q/431615/31097. Please do not post the same question on multiple sites. Each community should have an honest shot at answering without anybody's time being wasted. – D.W. Apr 10 at 0:56

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:

$$ABC$$

• Thanks a lot! This was very helpful! – Jarvis Ferns Apr 9 at 20:30
• @JarvisFerns you're welcome! :) – Bram28 Apr 9 at 20:39