Simplifying a 4-term equation using boolean algebra

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SUMMARY

The discussion focuses on simplifying a complex boolean equation using boolean algebra techniques. The original equation is simplified using complement and identity laws, resulting in a partially simplified form. The final simplified expression, derived using Karnaugh maps, is ~A * B * C + B * C * ~D + A * ~C * D + A * ~B * D. A key hint provided for further simplification is to factor out B from the expression.

PREREQUISITES
  • Understanding of boolean algebra concepts, including AND, OR, and NOT operations.
  • Familiarity with complement and identity laws in boolean algebra.
  • Knowledge of Karnaugh maps for simplifying boolean expressions.
  • Ability to perform factoring in algebraic expressions.
NEXT STEPS
  • Study boolean algebra factoring techniques to simplify expressions effectively.
  • Learn how to construct and utilize Karnaugh maps for boolean simplification.
  • Explore advanced boolean algebra laws, such as consensus and absorption laws.
  • Practice simplifying complex boolean expressions with multiple variables.
USEFUL FOR

This discussion is beneficial for students and professionals in electrical engineering, computer science, and anyone involved in digital logic design or boolean algebra simplification.

TheArtOfFail
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So, I have an equation:

~A * B * C * ~D + ~A * B * C * D + A * ~B * ~C * D + A * ~B * C * D + A * B * ~C * D + A * B * C * ~D

where * represents "AND" and + represents "OR", ~ being NOT.

Part of the reason I'm having trouble is due to the length of the equation.

So far, I've managed to use complement and identity laws to cancel out one pair of Cs and Ds, and ended up with this:

~A * B * C + A * ~B * D + A * B * ~C * D + A * B * C * ~D

Now, I know, through using Karnaugh maps, that the answer is this:

~A * B * C + B* C* ~D + A * ~C * D + A * ~B *D.

So, I'm close. I have two of the parameters. But I need to simplify two others.

I'm not asking for the straight answer, but can I have a hint as to what I need to do next?
 
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##(aBCd + aBCD) + (AbcD + AbCD) + ABcD + ABCd##
## = aBC + AbD + ABcD + ABCd##

Hint: Factor out B.
 

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