The following is the logic circuit:

enter image description here

I have to simplify the following:






  • 1
    $\begingroup$ 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

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 $A'+B'$:




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


  • $\begingroup$ 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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