Finding the complement using demorgans and involution (boolean alg)

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SUMMARY

The discussion focuses on using DeMorgan's relationships and the Involution law to find the complements of the Boolean function f(A,B,C,D) = [A+(BCD)'][(AD)'+B(C'+A)]. The user correctly applies DeMorgan's theorem and simplifies the expression to A'BCD + (AD)(B' + CA'). The conversation highlights the importance of recognizing simplifications such as AA' = 0 and A + A' = 1 in Boolean algebra. Additionally, constructing a Truth Table is recommended for verifying the correctness of the derived expression.

PREREQUISITES
  • Understanding of Boolean algebra concepts
  • Familiarity with DeMorgan's Theorems
  • Knowledge of the Involution law in Boolean functions
  • Ability to construct and analyze Truth Tables
NEXT STEPS
  • Study advanced applications of DeMorgan's Theorem in digital circuit design
  • Learn how to construct and interpret Truth Tables for complex Boolean expressions
  • Explore simplification techniques in Boolean algebra, including consensus and absorption laws
  • Investigate software tools for Boolean function minimization, such as Karnaugh maps or Quine-McCluskey algorithm
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Students and professionals in electrical engineering, computer science, or anyone involved in digital logic design and Boolean algebra simplification.

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Homework Statement


Use only DeMorgan's relationships and Involution to find the complements of the following functions:
a.) f(A,B,C,D) = [A+(BCD)'][(AD)'+B(C'+A)]


Homework Equations


Demorgans (x1 + x2 + ... + xn)' = x1'x2'...xn'

Involution (x')' = x

The Attempt at a Solution



[[A+(BCD)'][(AD)'+B(C'+A)]]' to find the compliment, then using demorgans
[A+(BCD)']' + [(AD)'+B(C'+A)]'
[A'(BCD)] + (AD)[B(C'+A)]'
A'BCD + (AD)[B' + (C'+A)']
A'BCD + (AD)(B' + CA')

from here I don't know where to go, i would think the right side of the equation could turn to ADB' + ADCA' but I'm not sure, if it can ADCA' would just be 0 since AA' = 0. Don't know if I can do that though, just looking for some input and hopefully I didn't make a mistake towards the begining.
 
Last edited:
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Hi buddyblakester, http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

Homework Equations


Demorgans (x1 + x2 + ... + xn) = x1'x2'...xn'
That is not a correct expression for De Morgan's theorem.
 
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had it on my paper right but yea typed it in wrong, thanks
 
I hadn't noticed it was just a typo.

i would think the right side of the equation could turn to ADB' + ADCA' but I'm not sure, if it can ADCA' would just be 0 since AA' = 0.
Yes, that looks right.

You can check by constructing a Truth Table for the original expression and for your answer.
 
ok cool, seems like AA' = 0 and A + A' = 1 can really reduce some of these kinds of equations in my homework. thanks for the feedback
 

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