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Simplify the following expressions to the simplest expression using De Morgan's theorem and Boolean algebra.

ABC+A'CD+B'CD

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

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

=(AB+(AB)'D)C

can anyone simplify it further and explain how you got there.Thanks in advance.

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    $\begingroup$ We have $AB\lor ABD=AB$ (absorption), and $ABD\lor (AB)'D=D$. So $AB\lor (AB)'D=AB\lor D$. $\endgroup$ – user10354138 Jun 3 at 8:49
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It always helps to expand the terms so they include all variables .. and then reorganize, and recombine in a more efficient way, adding or removing duplicates as needed.

The key principle is:

Adjacency

$P = PQ + PQ'$

Starting from your second expression:

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

$(ABD+ABD'+A'BD+A'B'D+AB'D+A'B'D)C=$

$(ABD+ABD'+(AB+A'B+AB'+A'B')D)C=$

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

$(AB+D)C$

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  • $\begingroup$ Thanks a lot!This was very helpful! $\endgroup$ – Jarvis Ferns Jun 3 at 12:04
  • $\begingroup$ @JarvisFerns You're welcome! :) $\endgroup$ – Bram28 Jun 3 at 12:28

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