Can F1 Expression Be Further Simplified?

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The discussion focuses on the simplification of the Boolean expression F1, initially presented as F1=(A’∙B∙C)+(A∙B’∙C’)+(A∙B∙C’)+(A∙B∙C). Participants explore whether this expression can be simplified further, leading to a breakdown of terms and the application of Boolean algebra rules. The expression is reordered and simplified to (1)BC + AC'(1), indicating that F1 can indeed be simplified. The conversation emphasizes the importance of understanding the steps involved in the simplification process. Overall, the simplification demonstrates the effectiveness of Boolean algebra in reducing complex expressions.
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Is it possible to simplify this further?

F1=(A’∙B∙C)+(A∙B’∙C’)+(A∙B∙C’)+(A∙B∙C)
 
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(a + -(bc) ) + ab * (c + -c) = f1 ??
 
bc + ac'
 
could you explain how you go that? Thanks so much!
 
mgord009 said:
Is it possible to simplify this further?

F1=(A’∙B∙C)+(A∙B’∙C’)+(A∙B∙C’)+(A∙B∙C)

= A'BC + ABC + AB'C' + ABC' by reordering the terms

= (A' + A) BC + AC'(B' + B)

= (1)BC + AC'(1)
 
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