Help Me simplify: C*(A+B) + ~A*B
2
$begingroup$
I know the answer is AC + ~AB, but how?
I have tried:
B*(~A+C) + A*C and stop.
Also, I have tried:
AC + BC + ~AB and have stopped.
There seems nowhere to go in either case.
digital-logic electrical
edited 9 hours ago
pipe
10.1k42656
asked 11 hours ago
Alan KazemianAlan Kazemian
183
New contributor
Alan Kazemian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
2
$begingroup$
I know the answer is AC + ~AB, but how?
I have tried:
B*(~A+C) + A*C and stop.
Also, I have tried:
AC + BC + ~AB and have stopped.
There seems nowhere to go in either case.
digital-logic electrical
edited 9 hours ago
pipe
10.1k42656
asked 11 hours ago
Alan KazemianAlan Kazemian
183
New contributor
Alan Kazemian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
4
$begingroup$
Show what you've tried and be explicit with where you are confused. Help us to help you.
$endgroup$
– Bort
10 hours ago
add a comment |
2
2
2
$begingroup$
I know the answer is AC + ~AB, but how?
I have tried:
B*(~A+C) + A*C and stop.
Also, I have tried:
AC + BC + ~AB and have stopped.
There seems nowhere to go in either case.
digital-logic electrical
edited 9 hours ago
pipe
10.1k42656
asked 11 hours ago
Alan KazemianAlan Kazemian
183
New contributor
Alan Kazemian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I know the answer is AC + ~AB, but how?
I have tried:
B*(~A+C) + A*C and stop.
Also, I have tried:
AC + BC + ~AB and have stopped.
There seems nowhere to go in either case.
digital-logic electrical
digital-logic electrical
edited 9 hours ago
pipe
10.1k42656
asked 11 hours ago
Alan KazemianAlan Kazemian
183
New contributor
Alan Kazemian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 9 hours ago
pipe
10.1k42656
asked 11 hours ago
Alan KazemianAlan Kazemian
183
New contributor
Alan Kazemian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 9 hours ago
pipe
10.1k42656
edited 9 hours ago
pipe
10.1k42656
edited 9 hours ago
pipe
10.1k42656
10.1k42656
asked 11 hours ago
Alan KazemianAlan Kazemian
183
New contributor
Alan Kazemian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 11 hours ago
Alan KazemianAlan Kazemian
183
asked 11 hours ago
Alan KazemianAlan Kazemian
183
183
New contributor
Alan Kazemian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Alan Kazemian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Alan Kazemian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
4
$begingroup$
Show what you've tried and be explicit with where you are confused. Help us to help you.
$endgroup$
– Bort
10 hours ago
add a comment |
4
$begingroup$
Show what you've tried and be explicit with where you are confused. Help us to help you.
$endgroup$
– Bort
10 hours ago
4
4
$begingroup$
Show what you've tried and be explicit with where you are confused. Help us to help you.
$endgroup$
– Bort
10 hours ago
$begingroup$
Show what you've tried and be explicit with where you are confused. Help us to help you.
$endgroup$
– Bort
10 hours ago
add a comment |
1 Answer
1
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8
$begingroup$
I will not give you the full solution, but the required non-trivial sidestep. You have got this far:
C(A + B) + A'B = AC + BC + A'B =
Now here is the sidestep. We know that (A+A') is 1, so we can do:
= AC + (A + A')BC + A'B
From here you will need to expand it and use the "OR absorption law" twice.
answered 10 hours ago
Eugene Sh.Eugene Sh.
7,5681830
$endgroup$
$begingroup$
@AlanKazemian (Be sure and hit the check mark so you can close the question)
$endgroup$
– KingDuken
10 hours ago
$begingroup$
@Eugene Sh. Expanding the brackets was an obvious step but your "sidestep" was one that I had not considered. Very useful and I will bear this in mind for future tasks. A Karnaugh Map would also have provided a easy solution to the problem.
$endgroup$
– Pzy
8 hours ago
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
8
$begingroup$
I will not give you the full solution, but the required non-trivial sidestep. You have got this far:
C(A + B) + A'B = AC + BC + A'B =
Now here is the sidestep. We know that (A+A') is 1, so we can do:
= AC + (A + A')BC + A'B
From here you will need to expand it and use the "OR absorption law" twice.
answered 10 hours ago
Eugene Sh.Eugene Sh.
7,5681830
$endgroup$
$begingroup$
@AlanKazemian (Be sure and hit the check mark so you can close the question)
$endgroup$
– KingDuken
10 hours ago
$begingroup$
@Eugene Sh. Expanding the brackets was an obvious step but your "sidestep" was one that I had not considered. Very useful and I will bear this in mind for future tasks. A Karnaugh Map would also have provided a easy solution to the problem.
$endgroup$
– Pzy
8 hours ago
add a comment |
8
$begingroup$
I will not give you the full solution, but the required non-trivial sidestep. You have got this far:
C(A + B) + A'B = AC + BC + A'B =
Now here is the sidestep. We know that (A+A') is 1, so we can do:
= AC + (A + A')BC + A'B
From here you will need to expand it and use the "OR absorption law" twice.
answered 10 hours ago
Eugene Sh.Eugene Sh.
7,5681830
$endgroup$
$begingroup$
@AlanKazemian (Be sure and hit the check mark so you can close the question)
$endgroup$
– KingDuken
10 hours ago
$begingroup$
@Eugene Sh. Expanding the brackets was an obvious step but your "sidestep" was one that I had not considered. Very useful and I will bear this in mind for future tasks. A Karnaugh Map would also have provided a easy solution to the problem.
$endgroup$
– Pzy
8 hours ago
add a comment |
8
8
8
$begingroup$
I will not give you the full solution, but the required non-trivial sidestep. You have got this far:
C(A + B) + A'B = AC + BC + A'B =
Now here is the sidestep. We know that (A+A') is 1, so we can do:
= AC + (A + A')BC + A'B
From here you will need to expand it and use the "OR absorption law" twice.
answered 10 hours ago
Eugene Sh.Eugene Sh.
7,5681830
$endgroup$
I will not give you the full solution, but the required non-trivial sidestep. You have got this far:
C(A + B) + A'B = AC + BC + A'B =
Now here is the sidestep. We know that (A+A') is 1, so we can do:
= AC + (A + A')BC + A'B
From here you will need to expand it and use the "OR absorption law" twice.
answered 10 hours ago
Eugene Sh.Eugene Sh.
7,5681830
edited 8 hours ago
answered 10 hours ago
Eugene Sh.Eugene Sh.
7,5681830
answered 10 hours ago
Eugene Sh.Eugene Sh.
7,5681830
answered 10 hours ago
Eugene Sh.Eugene Sh.
7,5681830
7,5681830
$begingroup$
@AlanKazemian (Be sure and hit the check mark so you can close the question)
$endgroup$
– KingDuken
10 hours ago
$begingroup$
@Eugene Sh. Expanding the brackets was an obvious step but your "sidestep" was one that I had not considered. Very useful and I will bear this in mind for future tasks. A Karnaugh Map would also have provided a easy solution to the problem.
$endgroup$
– Pzy
8 hours ago
add a comment |
$begingroup$
@AlanKazemian (Be sure and hit the check mark so you can close the question)
$endgroup$
– KingDuken
10 hours ago
$begingroup$
@Eugene Sh. Expanding the brackets was an obvious step but your "sidestep" was one that I had not considered. Very useful and I will bear this in mind for future tasks. A Karnaugh Map would also have provided a easy solution to the problem.
$endgroup$
– Pzy
8 hours ago
$begingroup$
@AlanKazemian (Be sure and hit the check mark so you can close the question)
$endgroup$
– KingDuken
10 hours ago
$begingroup$
@AlanKazemian (Be sure and hit the check mark so you can close the question)
$endgroup$
– KingDuken
10 hours ago
$begingroup$
@Eugene Sh. Expanding the brackets was an obvious step but your "sidestep" was one that I had not considered. Very useful and I will bear this in mind for future tasks. A Karnaugh Map would also have provided a easy solution to the problem.
$endgroup$
– Pzy
8 hours ago
$begingroup$
@Eugene Sh. Expanding the brackets was an obvious step but your "sidestep" was one that I had not considered. Very useful and I will bear this in mind for future tasks. A Karnaugh Map would also have provided a easy solution to the problem.
$endgroup$
– Pzy
8 hours ago
add a comment |
Alan Kazemian is a new contributor. Be nice, and check out our Code of Conduct.
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4
$begingroup$
Show what you've tried and be explicit with where you are confused. Help us to help you.
$endgroup$
– Bort
10 hours ago