Can the Boolean expression F1=(A'BC)+(AB'C')+(ABC')+(ABC) be simplified further?

[mgord009]

Is it possible to simplify this further?

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

 

[mgord009]

(a + -(bc) ) + ab * (c + -c) = f1 ??

 

[Stephen Tashi]

bc + ac'

 

[mgord009]

could you explain how you go that? Thanks so much!

 

[Stephen Tashi]

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)