How to simplify this XNOR expression? [closed]
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**Show that AB' + C(A'B' + AB) = AB' + AC + B'C**
I understand that the expression in brackets is the same as A XNOR B, but have no idea how to proceed towards the final answer.
logic
asked Dec 24 '18 at 16:08
Wei LunWei Lun
31
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closed as off-topic by Namaste, Abcd, José Carlos Santos, Leucippus, Shailesh Dec 25 '18 at 3:35
This question appears to be off-topic. The users who voted to close gave this specific reason:
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If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
**Show that AB' + C(A'B' + AB) = AB' + AC + B'C**
I understand that the expression in brackets is the same as A XNOR B, but have no idea how to proceed towards the final answer.
logic
asked Dec 24 '18 at 16:08
Wei LunWei Lun
31
$endgroup$
closed as off-topic by Namaste, Abcd, José Carlos Santos, Leucippus, Shailesh Dec 25 '18 at 3:35
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, Abcd, José Carlos Santos, Leucippus, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
**Show that AB' + C(A'B' + AB) = AB' + AC + B'C**
I understand that the expression in brackets is the same as A XNOR B, but have no idea how to proceed towards the final answer.
logic
asked Dec 24 '18 at 16:08
Wei LunWei Lun
31
$endgroup$
**Show that AB' + C(A'B' + AB) = AB' + AC + B'C**
I understand that the expression in brackets is the same as A XNOR B, but have no idea how to proceed towards the final answer.
logic
logic
asked Dec 24 '18 at 16:08
Wei LunWei Lun
31
asked Dec 24 '18 at 16:08
Wei LunWei Lun
31
asked Dec 24 '18 at 16:08
Wei LunWei Lun
31
asked Dec 24 '18 at 16:08
Wei LunWei Lun
31
asked Dec 24 '18 at 16:08
Wei LunWei Lun
31
31
closed as off-topic by Namaste, Abcd, José Carlos Santos, Leucippus, Shailesh Dec 25 '18 at 3:35
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, Abcd, José Carlos Santos, Leucippus, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Namaste, Abcd, José Carlos Santos, Leucippus, Shailesh Dec 25 '18 at 3:35
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, Abcd, José Carlos Santos, Leucippus, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
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oldest
votes
$begingroup$
$AB'+C(A'B'+AB)=AB'+A'B'C+ABC\=B'(A+A'C)+ABC\=B'(A+C)+ABC (*)\=AB'+B'C+ABC\=AB'+C(B'+AB)\=AB'+C(B'+A) (*)$
$(*)$ Absorption Law: $X+X'Y=X+Y$
answered Dec 24 '18 at 16:16
Shubham JohriShubham Johri
5,204718
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$begingroup$
Thanks so much!
$endgroup$
– Wei Lun
Dec 25 '18 at 5:12
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$AB'+C(A'B'+AB)=AB'+A'B'C+ABC\=B'(A+A'C)+ABC\=B'(A+C)+ABC (*)\=AB'+B'C+ABC\=AB'+C(B'+AB)\=AB'+C(B'+A) (*)$
$(*)$ Absorption Law: $X+X'Y=X+Y$
answered Dec 24 '18 at 16:16
Shubham JohriShubham Johri
5,204718
$endgroup$
$begingroup$
Thanks so much!
$endgroup$
– Wei Lun
Dec 25 '18 at 5:12
add a comment |
$begingroup$
$AB'+C(A'B'+AB)=AB'+A'B'C+ABC\=B'(A+A'C)+ABC\=B'(A+C)+ABC (*)\=AB'+B'C+ABC\=AB'+C(B'+AB)\=AB'+C(B'+A) (*)$
$(*)$ Absorption Law: $X+X'Y=X+Y$
answered Dec 24 '18 at 16:16
Shubham JohriShubham Johri
5,204718
$endgroup$
$begingroup$
Thanks so much!
$endgroup$
– Wei Lun
Dec 25 '18 at 5:12
add a comment |
$begingroup$
$AB'+C(A'B'+AB)=AB'+A'B'C+ABC\=B'(A+A'C)+ABC\=B'(A+C)+ABC (*)\=AB'+B'C+ABC\=AB'+C(B'+AB)\=AB'+C(B'+A) (*)$
$(*)$ Absorption Law: $X+X'Y=X+Y$
answered Dec 24 '18 at 16:16
Shubham JohriShubham Johri
5,204718
$endgroup$
$AB'+C(A'B'+AB)=AB'+A'B'C+ABC\=B'(A+A'C)+ABC\=B'(A+C)+ABC (*)\=AB'+B'C+ABC\=AB'+C(B'+AB)\=AB'+C(B'+A) (*)$
$(*)$ Absorption Law: $X+X'Y=X+Y$
answered Dec 24 '18 at 16:16
Shubham JohriShubham Johri
5,204718
answered Dec 24 '18 at 16:16
Shubham JohriShubham Johri
5,204718
answered Dec 24 '18 at 16:16
Shubham JohriShubham Johri
5,204718
answered Dec 24 '18 at 16:16
Shubham JohriShubham Johri
5,204718
5,204718
$begingroup$
Thanks so much!
$endgroup$
– Wei Lun
Dec 25 '18 at 5:12
add a comment |
$begingroup$
Thanks so much!
$endgroup$
– Wei Lun
Dec 25 '18 at 5:12
$begingroup$
Thanks so much!
$endgroup$
– Wei Lun
Dec 25 '18 at 5:12
$begingroup$
Thanks so much!
$endgroup$
– Wei Lun
Dec 25 '18 at 5:12
add a comment |
