Simplification of: AB + A'C + BC in boolean algebra
$begingroup$
I am trying to understand the simplification of the boolean expression:
AB + A'C + BC
I know it simplifies to
A'C + BC
And I understand why, but I cannot figure out how to perform the simplification through the expression using the boolean algebra identities. I was wondering if someone could show me the steps needed to do this. Thank you in advance.
boolean-algebra
asked Sep 5 '16 at 18:39
M. S.M. S.
1112
$endgroup$
add a comment |
$begingroup$
I am trying to understand the simplification of the boolean expression:
AB + A'C + BC
I know it simplifies to
A'C + BC
And I understand why, but I cannot figure out how to perform the simplification through the expression using the boolean algebra identities. I was wondering if someone could show me the steps needed to do this. Thank you in advance.
boolean-algebra
asked Sep 5 '16 at 18:39
M. S.M. S.
1112
$endgroup$
$begingroup$
If you get stuck in the future, try Karnaugh maps. They can help to simplify complicated boolean expressions.
$endgroup$
– Cameron Williams
Sep 5 '16 at 19:04
$begingroup$
Thank you for the suggestion and I know how to use them I just still don't know how to come to the answer with equation simplification.
$endgroup$
– M. S.
Sep 5 '16 at 22:12
add a comment |
$begingroup$
I am trying to understand the simplification of the boolean expression:
AB + A'C + BC
I know it simplifies to
A'C + BC
And I understand why, but I cannot figure out how to perform the simplification through the expression using the boolean algebra identities. I was wondering if someone could show me the steps needed to do this. Thank you in advance.
boolean-algebra
asked Sep 5 '16 at 18:39
M. S.M. S.
1112
$endgroup$
I am trying to understand the simplification of the boolean expression:
AB + A'C + BC
I know it simplifies to
A'C + BC
And I understand why, but I cannot figure out how to perform the simplification through the expression using the boolean algebra identities. I was wondering if someone could show me the steps needed to do this. Thank you in advance.
boolean-algebra
boolean-algebra
asked Sep 5 '16 at 18:39
M. S.M. S.
1112
asked Sep 5 '16 at 18:39
M. S.M. S.
1112
asked Sep 5 '16 at 18:39
M. S.M. S.
1112
asked Sep 5 '16 at 18:39
M. S.M. S.
1112
asked Sep 5 '16 at 18:39
M. S.M. S.
1112
1112
$begingroup$
If you get stuck in the future, try Karnaugh maps. They can help to simplify complicated boolean expressions.
$endgroup$
– Cameron Williams
Sep 5 '16 at 19:04
$begingroup$
Thank you for the suggestion and I know how to use them I just still don't know how to come to the answer with equation simplification.
$endgroup$
– M. S.
Sep 5 '16 at 22:12
add a comment |
$begingroup$
If you get stuck in the future, try Karnaugh maps. They can help to simplify complicated boolean expressions.
$endgroup$
– Cameron Williams
Sep 5 '16 at 19:04
$begingroup$
Thank you for the suggestion and I know how to use them I just still don't know how to come to the answer with equation simplification.
$endgroup$
– M. S.
Sep 5 '16 at 22:12
$begingroup$
If you get stuck in the future, try Karnaugh maps. They can help to simplify complicated boolean expressions.
$endgroup$
– Cameron Williams
Sep 5 '16 at 19:04
$begingroup$
If you get stuck in the future, try Karnaugh maps. They can help to simplify complicated boolean expressions.
$endgroup$
– Cameron Williams
Sep 5 '16 at 19:04
$begingroup$
Thank you for the suggestion and I know how to use them I just still don't know how to come to the answer with equation simplification.
$endgroup$
– M. S.
Sep 5 '16 at 22:12
$begingroup$
Thank you for the suggestion and I know how to use them I just still don't know how to come to the answer with equation simplification.
$endgroup$
– M. S.
Sep 5 '16 at 22:12
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
The two expressions are not equal. The first expression is true when A and B is true and C false but the second is false in this case.
answered Sep 5 '16 at 19:02
laissez_fairelaissez_faire
384
$endgroup$
$begingroup$
I might be mistaken but I believe that both equations return true in that instance. Because if A is true and C is false then A'C is true. Meaning the whole second equation is true because it is an OR.
$endgroup$
– M. S.
Sep 5 '16 at 22:05
$begingroup$
just for clarification because my notation isn't that common the two equations are (AB) + (A*(¬C)) + (BC) and the other one is (A*(¬C)) + (B*C)
$endgroup$
– M. S.
Sep 5 '16 at 22:08
$begingroup$
Okay, yes then I understand. I thought you had negation on the variable to the left of your prime character. One way to arrive at the simplified expression is: $AB+A(neg C)+BC=AB(C+(neg C))+A(neg C)(B+(neg B))+BC(A+(neg A))=ABC+AB(neg C)+AB(neg C)+A(neg B)(neg C)+ABC+(neg A)BC=ABC+AB(neg C)+A(neg B)(neg C)+(neg A)BC=BC(A+(neg A))+A(neg C)(B+(neg B))=BC+A(neg C)$
$endgroup$
– laissez_faire
Sep 6 '16 at 19:15
$begingroup$
I know this is late, but thank you very much!
$endgroup$
– M. S.
Nov 15 '16 at 2:59
add a comment |
$begingroup$
begin{align*}
&mathrel{phantom{=}}AB+A'C+BC\
&=AB+A'C+BC(A+A') quad text{($A+A'=1$, Complementarity law)}\
&=AB+A'C+ABC+A'BC\
&=AB+ABC+A'C+ABC quad text{(Associative law)}\
&=AB+A'C quad text{(Absorption law)}
end{align*}
edited Feb 12 '18 at 4:15
Saad
19.7k92352
answered Feb 12 '18 at 3:52
AnirbanAnirban
111
$endgroup$
add a comment |
$begingroup$
In this way , this can be simplified
LHS = AB+A'C+BC
= AB+A'C+BC (A+A') [ A+A'=1 ]
= AB+A'C+ABC+A'BC
= AB+ABC+A'C+A'BC
= AB (1+C)+A'C (1+B)
= AB+A'C [ 1+C=1 ]
=RHS..
answered Aug 3 '17 at 12:58
Priya yadavPriya yadav
1
$endgroup$
$begingroup$
Please use Math Jax in order to produce beautiful and beautifully formatted mathematical texts.
$endgroup$
– Daniele Tampieri
Jan 22 at 21:28
add a comment |
$begingroup$
a.b+a'.c+b.c
a.b+a'.c+b.c(a+a') {Complementary Law}
a.b+a'.c+(a.b.c+a'.b.c)
(a.b+a.b.c)+(a'.c+a'.c.b)
a.b(1+c)+a'.c(1+b) {As 1+c=1 and 1+b=1}
a.b+a'.c
edited Jul 8 '18 at 10:00
Martin Sleziak
44.8k10119272
answered Jul 8 '18 at 3:17
aritra dasaritra das
11
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The two expressions are not equal. The first expression is true when A and B is true and C false but the second is false in this case.
answered Sep 5 '16 at 19:02
laissez_fairelaissez_faire
384
$endgroup$
$begingroup$
I might be mistaken but I believe that both equations return true in that instance. Because if A is true and C is false then A'C is true. Meaning the whole second equation is true because it is an OR.
$endgroup$
– M. S.
Sep 5 '16 at 22:05
$begingroup$
just for clarification because my notation isn't that common the two equations are (AB) + (A*(¬C)) + (BC) and the other one is (A*(¬C)) + (B*C)
$endgroup$
– M. S.
Sep 5 '16 at 22:08
$begingroup$
Okay, yes then I understand. I thought you had negation on the variable to the left of your prime character. One way to arrive at the simplified expression is: $AB+A(neg C)+BC=AB(C+(neg C))+A(neg C)(B+(neg B))+BC(A+(neg A))=ABC+AB(neg C)+AB(neg C)+A(neg B)(neg C)+ABC+(neg A)BC=ABC+AB(neg C)+A(neg B)(neg C)+(neg A)BC=BC(A+(neg A))+A(neg C)(B+(neg B))=BC+A(neg C)$
$endgroup$
– laissez_faire
Sep 6 '16 at 19:15
$begingroup$
I know this is late, but thank you very much!
$endgroup$
– M. S.
Nov 15 '16 at 2:59
add a comment |
$begingroup$
The two expressions are not equal. The first expression is true when A and B is true and C false but the second is false in this case.
answered Sep 5 '16 at 19:02
laissez_fairelaissez_faire
384
$endgroup$
$begingroup$
I might be mistaken but I believe that both equations return true in that instance. Because if A is true and C is false then A'C is true. Meaning the whole second equation is true because it is an OR.
$endgroup$
– M. S.
Sep 5 '16 at 22:05
$begingroup$
just for clarification because my notation isn't that common the two equations are (AB) + (A*(¬C)) + (BC) and the other one is (A*(¬C)) + (B*C)
$endgroup$
– M. S.
Sep 5 '16 at 22:08
$begingroup$
Okay, yes then I understand. I thought you had negation on the variable to the left of your prime character. One way to arrive at the simplified expression is: $AB+A(neg C)+BC=AB(C+(neg C))+A(neg C)(B+(neg B))+BC(A+(neg A))=ABC+AB(neg C)+AB(neg C)+A(neg B)(neg C)+ABC+(neg A)BC=ABC+AB(neg C)+A(neg B)(neg C)+(neg A)BC=BC(A+(neg A))+A(neg C)(B+(neg B))=BC+A(neg C)$
$endgroup$
– laissez_faire
Sep 6 '16 at 19:15
$begingroup$
I know this is late, but thank you very much!
$endgroup$
– M. S.
Nov 15 '16 at 2:59
add a comment |
$begingroup$
The two expressions are not equal. The first expression is true when A and B is true and C false but the second is false in this case.
answered Sep 5 '16 at 19:02
laissez_fairelaissez_faire
384
$endgroup$
The two expressions are not equal. The first expression is true when A and B is true and C false but the second is false in this case.
answered Sep 5 '16 at 19:02
laissez_fairelaissez_faire
384
answered Sep 5 '16 at 19:02
laissez_fairelaissez_faire
384
answered Sep 5 '16 at 19:02
laissez_fairelaissez_faire
384
answered Sep 5 '16 at 19:02
laissez_fairelaissez_faire
384
384
$begingroup$
I might be mistaken but I believe that both equations return true in that instance. Because if A is true and C is false then A'C is true. Meaning the whole second equation is true because it is an OR.
$endgroup$
– M. S.
Sep 5 '16 at 22:05
$begingroup$
just for clarification because my notation isn't that common the two equations are (AB) + (A*(¬C)) + (BC) and the other one is (A*(¬C)) + (B*C)
$endgroup$
– M. S.
Sep 5 '16 at 22:08
$begingroup$
Okay, yes then I understand. I thought you had negation on the variable to the left of your prime character. One way to arrive at the simplified expression is: $AB+A(neg C)+BC=AB(C+(neg C))+A(neg C)(B+(neg B))+BC(A+(neg A))=ABC+AB(neg C)+AB(neg C)+A(neg B)(neg C)+ABC+(neg A)BC=ABC+AB(neg C)+A(neg B)(neg C)+(neg A)BC=BC(A+(neg A))+A(neg C)(B+(neg B))=BC+A(neg C)$
$endgroup$
– laissez_faire
Sep 6 '16 at 19:15
$begingroup$
I know this is late, but thank you very much!
$endgroup$
– M. S.
Nov 15 '16 at 2:59
add a comment |
$begingroup$
I might be mistaken but I believe that both equations return true in that instance. Because if A is true and C is false then A'C is true. Meaning the whole second equation is true because it is an OR.
$endgroup$
– M. S.
Sep 5 '16 at 22:05
$begingroup$
just for clarification because my notation isn't that common the two equations are (AB) + (A*(¬C)) + (BC) and the other one is (A*(¬C)) + (B*C)
$endgroup$
– M. S.
Sep 5 '16 at 22:08
$begingroup$
Okay, yes then I understand. I thought you had negation on the variable to the left of your prime character. One way to arrive at the simplified expression is: $AB+A(neg C)+BC=AB(C+(neg C))+A(neg C)(B+(neg B))+BC(A+(neg A))=ABC+AB(neg C)+AB(neg C)+A(neg B)(neg C)+ABC+(neg A)BC=ABC+AB(neg C)+A(neg B)(neg C)+(neg A)BC=BC(A+(neg A))+A(neg C)(B+(neg B))=BC+A(neg C)$
$endgroup$
– laissez_faire
Sep 6 '16 at 19:15
$begingroup$
I know this is late, but thank you very much!
$endgroup$
– M. S.
Nov 15 '16 at 2:59
$begingroup$
I might be mistaken but I believe that both equations return true in that instance. Because if A is true and C is false then A'C is true. Meaning the whole second equation is true because it is an OR.
$endgroup$
– M. S.
Sep 5 '16 at 22:05
$begingroup$
I might be mistaken but I believe that both equations return true in that instance. Because if A is true and C is false then A'C is true. Meaning the whole second equation is true because it is an OR.
$endgroup$
– M. S.
Sep 5 '16 at 22:05
$begingroup$
just for clarification because my notation isn't that common the two equations are (AB) + (A*(¬C)) + (BC) and the other one is (A*(¬C)) + (B*C)
$endgroup$
– M. S.
Sep 5 '16 at 22:08
$begingroup$
just for clarification because my notation isn't that common the two equations are (AB) + (A*(¬C)) + (BC) and the other one is (A*(¬C)) + (B*C)
$endgroup$
– M. S.
Sep 5 '16 at 22:08
$begingroup$
Okay, yes then I understand. I thought you had negation on the variable to the left of your prime character. One way to arrive at the simplified expression is: $AB+A(neg C)+BC=AB(C+(neg C))+A(neg C)(B+(neg B))+BC(A+(neg A))=ABC+AB(neg C)+AB(neg C)+A(neg B)(neg C)+ABC+(neg A)BC=ABC+AB(neg C)+A(neg B)(neg C)+(neg A)BC=BC(A+(neg A))+A(neg C)(B+(neg B))=BC+A(neg C)$
$endgroup$
– laissez_faire
Sep 6 '16 at 19:15
$begingroup$
Okay, yes then I understand. I thought you had negation on the variable to the left of your prime character. One way to arrive at the simplified expression is: $AB+A(neg C)+BC=AB(C+(neg C))+A(neg C)(B+(neg B))+BC(A+(neg A))=ABC+AB(neg C)+AB(neg C)+A(neg B)(neg C)+ABC+(neg A)BC=ABC+AB(neg C)+A(neg B)(neg C)+(neg A)BC=BC(A+(neg A))+A(neg C)(B+(neg B))=BC+A(neg C)$
$endgroup$
– laissez_faire
Sep 6 '16 at 19:15
$begingroup$
I know this is late, but thank you very much!
$endgroup$
– M. S.
Nov 15 '16 at 2:59
$begingroup$
I know this is late, but thank you very much!
$endgroup$
– M. S.
Nov 15 '16 at 2:59
add a comment |
$begingroup$
begin{align*}
&mathrel{phantom{=}}AB+A'C+BC\
&=AB+A'C+BC(A+A') quad text{($A+A'=1$, Complementarity law)}\
&=AB+A'C+ABC+A'BC\
&=AB+ABC+A'C+ABC quad text{(Associative law)}\
&=AB+A'C quad text{(Absorption law)}
end{align*}
edited Feb 12 '18 at 4:15
Saad
19.7k92352
answered Feb 12 '18 at 3:52
AnirbanAnirban
111
$endgroup$
add a comment |
$begingroup$
begin{align*}
&mathrel{phantom{=}}AB+A'C+BC\
&=AB+A'C+BC(A+A') quad text{($A+A'=1$, Complementarity law)}\
&=AB+A'C+ABC+A'BC\
&=AB+ABC+A'C+ABC quad text{(Associative law)}\
&=AB+A'C quad text{(Absorption law)}
end{align*}
edited Feb 12 '18 at 4:15
Saad
19.7k92352
answered Feb 12 '18 at 3:52
AnirbanAnirban
111
$endgroup$
add a comment |
$begingroup$
begin{align*}
&mathrel{phantom{=}}AB+A'C+BC\
&=AB+A'C+BC(A+A') quad text{($A+A'=1$, Complementarity law)}\
&=AB+A'C+ABC+A'BC\
&=AB+ABC+A'C+ABC quad text{(Associative law)}\
&=AB+A'C quad text{(Absorption law)}
end{align*}
edited Feb 12 '18 at 4:15
Saad
19.7k92352
answered Feb 12 '18 at 3:52
AnirbanAnirban
111
$endgroup$
begin{align*}
&mathrel{phantom{=}}AB+A'C+BC\
&=AB+A'C+BC(A+A') quad text{($A+A'=1$, Complementarity law)}\
&=AB+A'C+ABC+A'BC\
&=AB+ABC+A'C+ABC quad text{(Associative law)}\
&=AB+A'C quad text{(Absorption law)}
end{align*}
edited Feb 12 '18 at 4:15
Saad
19.7k92352
answered Feb 12 '18 at 3:52
AnirbanAnirban
111
edited Feb 12 '18 at 4:15
Saad
19.7k92352
edited Feb 12 '18 at 4:15
Saad
19.7k92352
edited Feb 12 '18 at 4:15
Saad
19.7k92352
19.7k92352
answered Feb 12 '18 at 3:52
AnirbanAnirban
111
answered Feb 12 '18 at 3:52
AnirbanAnirban
111
answered Feb 12 '18 at 3:52
AnirbanAnirban
111
111
add a comment |
add a comment |
$begingroup$
In this way , this can be simplified
LHS = AB+A'C+BC
= AB+A'C+BC (A+A') [ A+A'=1 ]
= AB+A'C+ABC+A'BC
= AB+ABC+A'C+A'BC
= AB (1+C)+A'C (1+B)
= AB+A'C [ 1+C=1 ]
=RHS..
answered Aug 3 '17 at 12:58
Priya yadavPriya yadav
1
$endgroup$
$begingroup$
Please use Math Jax in order to produce beautiful and beautifully formatted mathematical texts.
$endgroup$
– Daniele Tampieri
Jan 22 at 21:28
add a comment |
$begingroup$
In this way , this can be simplified
LHS = AB+A'C+BC
= AB+A'C+BC (A+A') [ A+A'=1 ]
= AB+A'C+ABC+A'BC
= AB+ABC+A'C+A'BC
= AB (1+C)+A'C (1+B)
= AB+A'C [ 1+C=1 ]
=RHS..
answered Aug 3 '17 at 12:58
Priya yadavPriya yadav
1
$endgroup$
$begingroup$
Please use Math Jax in order to produce beautiful and beautifully formatted mathematical texts.
$endgroup$
– Daniele Tampieri
Jan 22 at 21:28
add a comment |
$begingroup$
In this way , this can be simplified
LHS = AB+A'C+BC
= AB+A'C+BC (A+A') [ A+A'=1 ]
= AB+A'C+ABC+A'BC
= AB+ABC+A'C+A'BC
= AB (1+C)+A'C (1+B)
= AB+A'C [ 1+C=1 ]
=RHS..
answered Aug 3 '17 at 12:58
Priya yadavPriya yadav
1
$endgroup$
In this way , this can be simplified
LHS = AB+A'C+BC
= AB+A'C+BC (A+A') [ A+A'=1 ]
= AB+A'C+ABC+A'BC
= AB+ABC+A'C+A'BC
= AB (1+C)+A'C (1+B)
= AB+A'C [ 1+C=1 ]
=RHS..
answered Aug 3 '17 at 12:58
Priya yadavPriya yadav
1
answered Aug 3 '17 at 12:58
Priya yadavPriya yadav
1
answered Aug 3 '17 at 12:58
Priya yadavPriya yadav
1
answered Aug 3 '17 at 12:58
Priya yadavPriya yadav
1
1
$begingroup$
Please use Math Jax in order to produce beautiful and beautifully formatted mathematical texts.
$endgroup$
– Daniele Tampieri
Jan 22 at 21:28
add a comment |
$begingroup$
Please use Math Jax in order to produce beautiful and beautifully formatted mathematical texts.
$endgroup$
– Daniele Tampieri
Jan 22 at 21:28
$begingroup$
Please use Math Jax in order to produce beautiful and beautifully formatted mathematical texts.
$endgroup$
– Daniele Tampieri
Jan 22 at 21:28
$begingroup$
Please use Math Jax in order to produce beautiful and beautifully formatted mathematical texts.
$endgroup$
– Daniele Tampieri
Jan 22 at 21:28
add a comment |
$begingroup$
a.b+a'.c+b.c
a.b+a'.c+b.c(a+a') {Complementary Law}
a.b+a'.c+(a.b.c+a'.b.c)
(a.b+a.b.c)+(a'.c+a'.c.b)
a.b(1+c)+a'.c(1+b) {As 1+c=1 and 1+b=1}
a.b+a'.c
edited Jul 8 '18 at 10:00
Martin Sleziak
44.8k10119272
answered Jul 8 '18 at 3:17
aritra dasaritra das
11
$endgroup$
add a comment |
$begingroup$
a.b+a'.c+b.c
a.b+a'.c+b.c(a+a') {Complementary Law}
a.b+a'.c+(a.b.c+a'.b.c)
(a.b+a.b.c)+(a'.c+a'.c.b)
a.b(1+c)+a'.c(1+b) {As 1+c=1 and 1+b=1}
a.b+a'.c
edited Jul 8 '18 at 10:00
Martin Sleziak
44.8k10119272
answered Jul 8 '18 at 3:17
aritra dasaritra das
11
$endgroup$
add a comment |
$begingroup$
a.b+a'.c+b.c
a.b+a'.c+b.c(a+a') {Complementary Law}
a.b+a'.c+(a.b.c+a'.b.c)
(a.b+a.b.c)+(a'.c+a'.c.b)
a.b(1+c)+a'.c(1+b) {As 1+c=1 and 1+b=1}
a.b+a'.c
edited Jul 8 '18 at 10:00
Martin Sleziak
44.8k10119272
answered Jul 8 '18 at 3:17
aritra dasaritra das
11
$endgroup$
a.b+a'.c+b.c
a.b+a'.c+b.c(a+a') {Complementary Law}
a.b+a'.c+(a.b.c+a'.b.c)
(a.b+a.b.c)+(a'.c+a'.c.b)
a.b(1+c)+a'.c(1+b) {As 1+c=1 and 1+b=1}
a.b+a'.c
edited Jul 8 '18 at 10:00
Martin Sleziak
44.8k10119272
answered Jul 8 '18 at 3:17
aritra dasaritra das
11
edited Jul 8 '18 at 10:00
Martin Sleziak
44.8k10119272
edited Jul 8 '18 at 10:00
Martin Sleziak
44.8k10119272
edited Jul 8 '18 at 10:00
Martin Sleziak
44.8k10119272
44.8k10119272
answered Jul 8 '18 at 3:17
aritra dasaritra das
11
answered Jul 8 '18 at 3:17
aritra dasaritra das
11
answered Jul 8 '18 at 3:17
aritra dasaritra das
11
11
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$begingroup$
If you get stuck in the future, try Karnaugh maps. They can help to simplify complicated boolean expressions.
$endgroup$
– Cameron Williams
Sep 5 '16 at 19:04
$begingroup$
Thank you for the suggestion and I know how to use them I just still don't know how to come to the answer with equation simplification.
$endgroup$
– M. S.
Sep 5 '16 at 22:12