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if (A,B,C)=AB+AC+BC then

F((A’,B’,C’).F(A’,B,C’).F(A,B’.C’))=?

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F(A,B,C)=AB+AC+BC

F(A',B',C')=A'B'+A'C'+B'C'

F(A',B,C')=A'B+A'C'+BC'

F(A,B',C')=AB'+AC'+B'C'


Now, F(A',B',C') . F(A',B,C') . F(A,B',C') = [A'B'+A'C'+B'C'] . [A'B+A'C'+BC'] . [AB'+AC'+B'C']

                                                               = [ A'B'C'+A'BC'+A'C' ] . [AB'+AC'+B'C']  (By Applying Distributive)

                                                               = [A'B'C'+A'C'] . [AB'+AC'+B'C'] (By Applying Absorption Law)

                                                               = A'B'C'

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