Answer to Question #105964 in Discrete Mathematics for yinusa kazeem

Question

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

Answer 1

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

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

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