Help with simplifying boolean expression

  • #1
< Mentor Note -- thread moved to HH from the technical math forums, so no HH Template is shown >

(A+B)&(C+D) + (A+B)&(C+D)' + C
(A+B)&(C+D) + (A+B)&(C'&D') + C by deMorgans
(A+B)&[(C+D)+(C'&D')] + C by Distributive

I'm just wondering if I did anything wrong in this simplification or if it can be simplified any further.
 
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Answers and Replies

  • #2
Svein
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(A+B)&(C+D) + (A+B)&(C+D)' + C
(A+B)&(C+D) + (A+B)&(C'&D') + C by deMorgans
(A+B)&[(C+D)+(C'&D')] + C by Distributive

I'm just wondering if I did anything wrong in this simplification or if it can be simplified any further.
Well, use the distributive law on the first part of first line - (A+B)&(C+D) + (A+B)&(C+D)' = (A+B)&((C+D) + (C+D)') = (A+B) (since P + P' = TRUE).
 
  • #3
Well, use the distributive law on the first part of first line - (A+B)&(C+D) + (A+B)&(C+D)' = (A+B)&((C+D) + (C+D)') = (A+B) (since P + P' = TRUE).
Thanks, I didn't even see that.
 

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