find the sum \\ (\\bar{AB}+\\bar{BC}+\\bar{CD}+\\bar{DE}+\\bar{EF}\\)

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$\overline{A B} + \overline{BC} + \overline{C D} + \overline{D E} + \overline{EF}$

De Morgan’s law

$\overline{XY}=\bar{X}+\bar{Y}$

Then

$\overline{AB}+\overline{BC}+\overline{CD}+\overline{DE}+\overline{EF}=$ $=\bar{A}+\bar{B}+\bar{B}+\bar{C}+\bar{C}+\bar{D}+\bar{D}+\bar{E}+\bar{E}+\bar{F}=$

Idempotent law

$X+X=X$

Hence

$\overline{AB}+\overline{BC}+\overline{CD}+\overline{DE}+\overline{EF}=$ $=\bar{A}+\bar{B}+\bar{B}+\bar{C}+\bar{C}+\bar{D}+\bar{D}+\bar{E}+\bar{E}+\bar{F}=$ $=\bar{A}+\bar{B}+\bar{C}+\bar{D}+\bar{E}+\bar{F}$

Use De Morgan’s law

$\bar{A}+\bar{B}+\bar{C}+\bar{D}+\bar{E}+\bar{F}=\overline{ABCDEF}$

Therefore

$\overline{AB}+\overline{BC}+\overline{CD}+\overline{DE}+\overline{EF}=\overline{ABCDEF}$

find the sum \\ (\\bar{AB}+\\bar{BC}+\\bar{CD}+\\bar{DE}+\\bar{EF}\\)

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Question ID: mtid-5-stid-8-sqid-4002-qpid-2701