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enter image description here

I have a question regarding simplifying a circuit of a function below that has 5 logic gates in original.

f = (A + B) * (C + D) + (A + B) * (C + D)' + C

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

= (A + B) * 1 + C Complement

= (A + B) + C

Now, I have reduced to 2 logic gates from 5. But, here, am I allowed to change (A + B) + C to A + B + C, so that I can reduce number of gates (i.e., 1 logic gate) even more? If I am allowed, is there a name for this process (what kind of law is this)?

Thank you in advance.

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    \$\begingroup\$ Previously asked at math.stackexchange.com/q/1126470/18398 \$\endgroup\$ – Joel Reyes Noche Jan 30 '15 at 13:04
  • \$\begingroup\$ You should always link to multiple instances of the same question whenever you ask the same question on different StackExchange sites. \$\endgroup\$ – Joel Reyes Noche Jan 30 '15 at 13:04
  • \$\begingroup\$ I assume that (A+B)*(C+D)' means ((A+B)*((C+D)'). Wouldn't your original need 6 logic gates? Did you count the NOT gate in (C+D)'? (Remember, you also have a (C+D) term.) \$\endgroup\$ – Joel Reyes Noche Jan 30 '15 at 13:08
  • \$\begingroup\$ Also, what is the word "Complement" in the third line about? It seems like it shouldn't be there. \$\endgroup\$ – Joel Reyes Noche Jan 30 '15 at 13:10
  • \$\begingroup\$ @JoelReyesNoche It's just a tag for complement law which states x+x'=1. \$\endgroup\$ – IANIK Jan 30 '15 at 13:21
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Yes, you can change the two 2-input OR gates to one 3-input OR gate. As for it being "allowed," that depends. Who is the one giving permission? If it's a teacher, then you should ask her/him what is "allowed." I am not familiar with any formal name for this substitution.

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    \$\begingroup\$ As was mentioned in the comments and an answer in the MSE version of this question, the name for this "process" is associativity. \$\endgroup\$ – Joel Reyes Noche Jan 30 '15 at 13:44
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You can also plot the equation using the technique called Karnaugh Mapping and visually see how your equation simplifies down. Look for an explanation of Karnaugh Mapping here:

http://www.facstaff.bucknell.edu/mastascu/elessonshtml/Logic/Logic3.html

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    \$\begingroup\$ We like answers here to at least be self-contained with the basic information. In this case, it would be good to give some idea what Karnaugh maps are, then you can use a link for more detail. \$\endgroup\$ – Olin Lathrop Jan 30 '15 at 16:39

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