Find the sum of AB+BC+CD+DE+EF
Answers
Answer:
De Morgan’s law
\overline{XY}=\bar{X}+\bar{Y}XY=Xˉ+Yˉ
Then
\overline{AB}+\overline{BC}+\overline{CD}+\overline{DE}+\overline{EF}=AB+BC+CD+DE+EF==\bar{A}+\bar{B}+\bar{B}+\bar{C}+\bar{C}+\bar{D}+\bar{D}+\bar{E}+\bar{E}+\bar{F}==Aˉ+Bˉ+Bˉ+Cˉ+Cˉ+Dˉ+Dˉ+Eˉ+Eˉ+Fˉ=
Idempotent law
X+X=XX+X=X
Hence
\overline{AB}+\overline{BC}+\overline{CD}+\overline{DE}+\overline{EF}=AB+BC+CD+DE+EF==\bar{A}+\bar{B}+\bar{B}+\bar{C}+\bar{C}+\bar{D}+\bar{D}+\bar{E}+\bar{E}+\bar{F}==Aˉ+Bˉ+Bˉ+Cˉ+Cˉ+Dˉ+Dˉ+Eˉ+Eˉ+Fˉ==\bar{A}+\bar{B}+\bar{C}+\bar{D}+\bar{E}+\bar{F}=Aˉ+Bˉ+Cˉ+Dˉ+Eˉ+Fˉ
Use De Morgan’s law
\bar{A}+\bar{B}+\bar{C}+\bar{D}+\bar{E}+\bar{F}=\overline{ABCDEF}Aˉ+Bˉ+Cˉ+Dˉ+Eˉ+Fˉ=ABCDEF
Therefore
\overline{AB}+\overline{BC}+\overline{CD}+\overline{DE}+\overline{EF}=\overline{ABCDEF}AB+BC+CD+DE+EF=ABCDEF