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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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closed as off-topic by Dave Tweed Apr 10 at 11:31

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    \$\begingroup\$ Draw a truth table for both - the original and the result. Compare. Note that the final expression can be further simplified. \$\endgroup\$ – Eugene Sh. Apr 9 at 15:28
  • \$\begingroup\$ Actually I'm given a logic circuit and have to simplify the expression in order to get the truth table. \$\endgroup\$ – Jarvis Ferns Apr 9 at 15:32
  • \$\begingroup\$ You can get a truth table from a non-simplified expression. This one is a bit complicated, but not a big deal. You can fill up half of it right away noting the AND with C of the whole expression. \$\endgroup\$ – Eugene Sh. Apr 9 at 15:34
  • \$\begingroup\$ Unless I'm reading the parenthesis wrong, the answer is just C. Everything is or'ed with that inner C + B, then and'd with C. Are you sure you copied it down correctly? \$\endgroup\$ – yhyrcanus Apr 9 at 19:02
  • \$\begingroup\$ @yhyrcanus I added the logic circuit and made some changes \$\endgroup\$ – Jarvis Ferns Apr 9 at 19:25
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You still messed up your initial statement.

((A NAND B) XOR (B OR C)) AND C

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

Note it's an AND between the ((AB)')' and the (B+C).

I hope you can get it from there. Still highly suggest making a truth table. It's super obvious what the answer is if you draw out the truth table.

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  • \$\begingroup\$ Thanks a lot! But I was unable to create a truth table first, so i finished the simplification. \$\endgroup\$ – Jarvis Ferns Apr 9 at 20:49
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(((AB)′)′(B+C)+(AB)′(B+C)′)C

=(AB(B+C)+(A′+B′)B′C′)C

=(AB+B′C′)C

=ABC+B′C′C

=ABC

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    \$\begingroup\$ Not mine, but I suspect it's the lack of words. Welcome to SE.EE too :) \$\endgroup\$ – awjlogan Apr 10 at 8:16

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