Boolean algebra simplification ?

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The discussion revolves around the simplification of Boolean algebra expressions, specifically using DeMorgan’s Theorems. A participant expresses uncertainty about their approach to a problem, questioning the presence of a '+' instead of a '*' in a specific line of their solution. They suggest an alternative direction for simplification, highlighting that the expression (¬AC)¬C can be simplified effectively. The conversation emphasizes the importance of understanding the application of Boolean rules for accurate simplification. Overall, the thread focuses on clarifying steps in Boolean algebra simplification techniques.
KaliBanda
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Homework Statement



t1eVU.png


Homework Equations



DeMorgan’s Theorems.

The Attempt at a Solution



xBT1D.png


I've had a go at it, not sure if I'm heading in the right direction though.

Thanks for any help.
 
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Where does the second + in line 4 come from? I would expect * there, as you had it in line 3.
Anyway, I would use another direction there:
##(\overline{AC})\overline{C}## has a nice simplification.
 

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