Answer:
The Following are the answer to the given statement:
Explanation:
Given value:
[tex]a) \ Y= \bar{A}BC+\bar{B\bar{C}} +BC \\\\b) \ Y= \bar{(A+B+C)}D+AD+B \\\\c) \ Y= ABCD+\bar{A} B \bar{C} D + \bar (\bar{B} + D) \ \ E[/tex]
In latex code, in typing there is some error so, please find the question.
Solve point a:
[tex]a) \ Y= \bar{A}BC+\bar{B\bar{C}} +BC \\\\[/tex]
[tex]= (\bar{A} + 1 ) BC+\bar{B\bar{C}} \\\\= (\bar{B} + C ) +BC\\\\= \bar{B} + C +BC\\\\= \bar{B} + C(B+1) \\\\= \bar{B} + C \\\\[/tex]
Solve point b:
[tex]b) \ Y= \bar{(A+B+C)}D+AD+B \\\\[/tex]
[tex]Y= \bar{A}+\bar{B}+\bar{C}D+AD+B \\\\[/tex]
[tex]= \bar{C}D+AD+B \\\\[/tex]
Solve point c:
[tex]c) \ Y= ABCD+\bar{A} B \bar{C} D + \bar (\bar{B} + D) \ \ E[/tex]
[tex]=ABCD+\bar{A}B\bar{C}D+ B\bar{D}E\\\\=(AC+\bar{A}\bar{C} )BD+ B\bar{D}E\\\\=(A \circ C)BD+ B\bar{D}E\\\\[/tex]
please find the circuits attachment for the above point.
