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The following is the logic circuit:

enter image description here

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

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    $\begingroup$ One quick way to check, to yourself, whether it's correct is to do a truth table. $\endgroup$ – Arthur Apr 9 at 18:55
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    $\begingroup$ How did you reach the second line? Where did $ABC$ come into the picture? $\endgroup$ – Shubham Johri Apr 9 at 19:14
  • $\begingroup$ 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. $\endgroup$ – Bram28 Apr 9 at 19:19
  • $\begingroup$ I made some changes $\endgroup$ – Jarvis Ferns Apr 9 at 19:20
  • $\begingroup$ 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. $\endgroup$ – D.W. Apr 10 at 0:56
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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:

$ABC$

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

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