Mixing
Boolean Algebra: Where did I go wrong with the simplification
Clash Royale CLAN TAG#URR8PPP
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Sorry, I don't know how to use the MathJax syntax. I am representing a not bar with
Formatted in (step used to solve : what I equated it to using that step)
Start: $Acdot overlineBcdot C+Ccdot Acdot B$
Demorgan: A•($overlineB$+$overlineC$)•($overlineC$+$overlineA$)•B
Distribution: A•($overlineC$+$overlineA$)•B•$overlineB$+A•($overlineC$+$overlineA$)•B•$overlineC$
A•$overlineA$=0 : 0+A•($overlineC$+$overlineA$)•B•$overlineC$
A+0=A : A•($overlineC$+$overlineA$)•B•$overlineC$
Distribution: A•B•$overlineC$•$overlineC$+A•B•$overlineC$•$overlineA$
A+A=A : A•B•$overlineC$•$overlineA$
A•$overlineA$=0 : 0
Final: 0
Sites are giving me this as the correct simplification: A•B•C
https://www.dcode.fr/boolean-expressions-calculator
boolean-algebra
edited Aug 2 at 16:37
Bram28
54.5k33880
asked Aug 2 at 16:16
John Smith
11
add a comment |Â
up vote
0
down vote
favorite
Sorry, I don't know how to use the MathJax syntax. I am representing a not bar with
Formatted in (step used to solve : what I equated it to using that step)
Start: $Acdot overlineBcdot C+Ccdot Acdot B$
Demorgan: A•($overlineB$+$overlineC$)•($overlineC$+$overlineA$)•B
Distribution: A•($overlineC$+$overlineA$)•B•$overlineB$+A•($overlineC$+$overlineA$)•B•$overlineC$
A•$overlineA$=0 : 0+A•($overlineC$+$overlineA$)•B•$overlineC$
A+0=A : A•($overlineC$+$overlineA$)•B•$overlineC$
Distribution: A•B•$overlineC$•$overlineC$+A•B•$overlineC$•$overlineA$
A+A=A : A•B•$overlineC$•$overlineA$
A•$overlineA$=0 : 0
Final: 0
Sites are giving me this as the correct simplification: A•B•C
https://www.dcode.fr/boolean-expressions-calculator
boolean-algebra
edited Aug 2 at 16:37
Bram28
54.5k33880
asked Aug 2 at 16:16
John Smith
11
Here a MathJax Tutorial
– saulspatz
Aug 2 at 16:17
"A+A=A : A•B•C•A." But you don't have "A•B•C•A+A•B•C•A" you have "A•B•C•C+A•B•C•A " and A•B•C•C$ne$A•B•C•A.
– fleablood
Aug 2 at 16:23
Thank you. I don't know why I thought that.
– John Smith
Aug 2 at 16:25
I edited and provided the proper Mathjax formula for your first expression ... can you edit and do the rest?
– Bram28
Aug 2 at 16:38
"I don't know why I thought that" A quasi-dyslexic proof reading over-sight. Frustratingly easy to make.
– fleablood
Aug 2 at 16:47
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Sorry, I don't know how to use the MathJax syntax. I am representing a not bar with
Formatted in (step used to solve : what I equated it to using that step)
Start: $Acdot overlineBcdot C+Ccdot Acdot B$
Demorgan: A•($overlineB$+$overlineC$)•($overlineC$+$overlineA$)•B
Distribution: A•($overlineC$+$overlineA$)•B•$overlineB$+A•($overlineC$+$overlineA$)•B•$overlineC$
A•$overlineA$=0 : 0+A•($overlineC$+$overlineA$)•B•$overlineC$
A+0=A : A•($overlineC$+$overlineA$)•B•$overlineC$
Distribution: A•B•$overlineC$•$overlineC$+A•B•$overlineC$•$overlineA$
A+A=A : A•B•$overlineC$•$overlineA$
A•$overlineA$=0 : 0
Final: 0
Sites are giving me this as the correct simplification: A•B•C
https://www.dcode.fr/boolean-expressions-calculator
boolean-algebra
edited Aug 2 at 16:37
Bram28
54.5k33880
asked Aug 2 at 16:16
John Smith
11
Sorry, I don't know how to use the MathJax syntax. I am representing a not bar with
Formatted in (step used to solve : what I equated it to using that step)
Start: $Acdot overlineBcdot C+Ccdot Acdot B$
Demorgan: A•($overlineB$+$overlineC$)•($overlineC$+$overlineA$)•B
Distribution: A•($overlineC$+$overlineA$)•B•$overlineB$+A•($overlineC$+$overlineA$)•B•$overlineC$
A•$overlineA$=0 : 0+A•($overlineC$+$overlineA$)•B•$overlineC$
A+0=A : A•($overlineC$+$overlineA$)•B•$overlineC$
Distribution: A•B•$overlineC$•$overlineC$+A•B•$overlineC$•$overlineA$
A+A=A : A•B•$overlineC$•$overlineA$
A•$overlineA$=0 : 0
Final: 0
Sites are giving me this as the correct simplification: A•B•C
https://www.dcode.fr/boolean-expressions-calculator
boolean-algebra
edited Aug 2 at 16:37
Bram28
54.5k33880
asked Aug 2 at 16:16
John Smith
11
edited Aug 2 at 16:37
Bram28
54.5k33880
edited Aug 2 at 16:37
Bram28
54.5k33880
edited Aug 2 at 16:37
Bram28
54.5k33880
54.5k33880
asked Aug 2 at 16:16
John Smith
11
asked Aug 2 at 16:16
John Smith
11
asked Aug 2 at 16:16
John Smith
11
11
Here a MathJax Tutorial
– saulspatz
Aug 2 at 16:17
"A+A=A : A•B•C•A." But you don't have "A•B•C•A+A•B•C•A" you have "A•B•C•C+A•B•C•A " and A•B•C•C$ne$A•B•C•A.
– fleablood
Aug 2 at 16:23
Thank you. I don't know why I thought that.
– John Smith
Aug 2 at 16:25
I edited and provided the proper Mathjax formula for your first expression ... can you edit and do the rest?
– Bram28
Aug 2 at 16:38
"I don't know why I thought that" A quasi-dyslexic proof reading over-sight. Frustratingly easy to make.
– fleablood
Aug 2 at 16:47
add a comment |Â
Here a MathJax Tutorial
– saulspatz
Aug 2 at 16:17
"A+A=A : A•B•C•A." But you don't have "A•B•C•A+A•B•C•A" you have "A•B•C•C+A•B•C•A " and A•B•C•C$ne$A•B•C•A.
– fleablood
Aug 2 at 16:23
Thank you. I don't know why I thought that.
– John Smith
Aug 2 at 16:25
I edited and provided the proper Mathjax formula for your first expression ... can you edit and do the rest?
– Bram28
Aug 2 at 16:38
"I don't know why I thought that" A quasi-dyslexic proof reading over-sight. Frustratingly easy to make.
– fleablood
Aug 2 at 16:47
Here a MathJax Tutorial
– saulspatz
Aug 2 at 16:17
Here a MathJax Tutorial
– saulspatz
Aug 2 at 16:17
"A+A=A : A•B•C•A." But you don't have "A•B•C•A+A•B•C•A" you have "A•B•C•C+A•B•C•A " and A•B•C•C$ne$A•B•C•A.
– fleablood
Aug 2 at 16:23
"A+A=A : A•B•C•A." But you don't have "A•B•C•A+A•B•C•A" you have "A•B•C•C+A•B•C•A " and A•B•C•C$ne$A•B•C•A.
– fleablood
Aug 2 at 16:23
Thank you. I don't know why I thought that.
– John Smith
Aug 2 at 16:25
Thank you. I don't know why I thought that.
– John Smith
Aug 2 at 16:25
I edited and provided the proper Mathjax formula for your first expression ... can you edit and do the rest?
– Bram28
Aug 2 at 16:38
I edited and provided the proper Mathjax formula for your first expression ... can you edit and do the rest?
– Bram28
Aug 2 at 16:38
"I don't know why I thought that" A quasi-dyslexic proof reading over-sight. Frustratingly easy to make.
– fleablood
Aug 2 at 16:47
"I don't know why I thought that" A quasi-dyslexic proof reading over-sight. Frustratingly easy to make.
– fleablood
Aug 2 at 16:47
add a comment |Â
3 Answers
3
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0
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Here is your mistake:
Distribution: A•B•$overlineC$•$overlineC$+A•B•$overlineC$•$overlineA$
A+A=A : A•B•$overlineC$•$overlineA$
No. The two terms are different. In fact, the second term contains $Acdot barA$, and thus works out to $0$, but the first term simplifies to just $Acdot B cdot barC$ ... as that website correctly finds.
answered Aug 2 at 16:35
Bram28
54.5k33880
add a comment |Â
up vote
0
down vote
You have line $6$
6) $A•B•overlineC•overlineC+A•B•overlineC•overlineA$
Then one line 7: you claim as $A+ A = A$ that $A•B•overlineC•overlineC+A•B•overlineC•overlineA= A•B•overlineC•overlineA$
But $A•B•overlineC•overlineC$ and $A•B•overlineC•overlineA$ are not the same.
Instead use $Acdot A = A$ to get that $overlineC•overlineC = overlineC$ to get:
7) $A•B•overlineC+A•B•overlineC•overlineA$
Then distribute:
8) $A•B•overlineC(1 + overline A)$
9) $1 + A = 1 : A•B•overlineCcdot 1$
10) $1cdot A = A: Acdot Bcdot overlineC$.
...or....
8) $Aoverline A = 0: Acdot Bcdot overlineC + 0$
9) $A + 0 = A: Acdot Bcdot overlineC$
answered Aug 2 at 16:45
fleablood
60.1k22575
add a comment |Â
up vote
0
down vote
Analogous with sets:
- $Acap((Bcap C)cup(Ccap A))^complementcap B$
- $Acap((Bcap C)^complementcap(Ccap A)^complement)cap B$
- $Acap((B^complementcup C^complement)cap(C^complementcup A^complement))cap B$
- $Acap((B^complementcap C^complement)cup(B^complementcap A^complement)cup (C^complementcap C^complement)cup(C^complementcap A^complement))cap B$
- $(Acap B^complementcap C^complementcap B)cup(Acap B^complementcap A^complementcap B)cup(Acap C^complementcap C^complementcap B)cup(Acap C^complementcap A^complementcap B)$
- $varnothingcupvarnothingcup(Acap C^complementcap C^complementcap B)cupvarnothing$
- $Acap C^complementcap B$
answered Aug 2 at 17:11
drhab
85.8k540118
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Here is your mistake:
Distribution: A•B•$overlineC$•$overlineC$+A•B•$overlineC$•$overlineA$
A+A=A : A•B•$overlineC$•$overlineA$
No. The two terms are different. In fact, the second term contains $Acdot barA$, and thus works out to $0$, but the first term simplifies to just $Acdot B cdot barC$ ... as that website correctly finds.
answered Aug 2 at 16:35
Bram28
54.5k33880
add a comment |Â
up vote
0
down vote
Here is your mistake:
Distribution: A•B•$overlineC$•$overlineC$+A•B•$overlineC$•$overlineA$
A+A=A : A•B•$overlineC$•$overlineA$
No. The two terms are different. In fact, the second term contains $Acdot barA$, and thus works out to $0$, but the first term simplifies to just $Acdot B cdot barC$ ... as that website correctly finds.
answered Aug 2 at 16:35
Bram28
54.5k33880
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Here is your mistake:
Distribution: A•B•$overlineC$•$overlineC$+A•B•$overlineC$•$overlineA$
A+A=A : A•B•$overlineC$•$overlineA$
No. The two terms are different. In fact, the second term contains $Acdot barA$, and thus works out to $0$, but the first term simplifies to just $Acdot B cdot barC$ ... as that website correctly finds.
answered Aug 2 at 16:35
Bram28
54.5k33880
Here is your mistake:
Distribution: A•B•$overlineC$•$overlineC$+A•B•$overlineC$•$overlineA$
A+A=A : A•B•$overlineC$•$overlineA$
No. The two terms are different. In fact, the second term contains $Acdot barA$, and thus works out to $0$, but the first term simplifies to just $Acdot B cdot barC$ ... as that website correctly finds.
answered Aug 2 at 16:35
Bram28
54.5k33880
answered Aug 2 at 16:35
Bram28
54.5k33880
answered Aug 2 at 16:35
Bram28
54.5k33880
answered Aug 2 at 16:35
Bram28
54.5k33880
54.5k33880
add a comment |Â
add a comment |Â
up vote
0
down vote
You have line $6$
6) $A•B•overlineC•overlineC+A•B•overlineC•overlineA$
Then one line 7: you claim as $A+ A = A$ that $A•B•overlineC•overlineC+A•B•overlineC•overlineA= A•B•overlineC•overlineA$
But $A•B•overlineC•overlineC$ and $A•B•overlineC•overlineA$ are not the same.
Instead use $Acdot A = A$ to get that $overlineC•overlineC = overlineC$ to get:
7) $A•B•overlineC+A•B•overlineC•overlineA$
Then distribute:
8) $A•B•overlineC(1 + overline A)$
9) $1 + A = 1 : A•B•overlineCcdot 1$
10) $1cdot A = A: Acdot Bcdot overlineC$.
...or....
8) $Aoverline A = 0: Acdot Bcdot overlineC + 0$
9) $A + 0 = A: Acdot Bcdot overlineC$
answered Aug 2 at 16:45
fleablood
60.1k22575
add a comment |Â
up vote
0
down vote
You have line $6$
6) $A•B•overlineC•overlineC+A•B•overlineC•overlineA$
Then one line 7: you claim as $A+ A = A$ that $A•B•overlineC•overlineC+A•B•overlineC•overlineA= A•B•overlineC•overlineA$
But $A•B•overlineC•overlineC$ and $A•B•overlineC•overlineA$ are not the same.
Instead use $Acdot A = A$ to get that $overlineC•overlineC = overlineC$ to get:
7) $A•B•overlineC+A•B•overlineC•overlineA$
Then distribute:
8) $A•B•overlineC(1 + overline A)$
9) $1 + A = 1 : A•B•overlineCcdot 1$
10) $1cdot A = A: Acdot Bcdot overlineC$.
...or....
8) $Aoverline A = 0: Acdot Bcdot overlineC + 0$
9) $A + 0 = A: Acdot Bcdot overlineC$
answered Aug 2 at 16:45
fleablood
60.1k22575
add a comment |Â
up vote
0
down vote
up vote
0
down vote
You have line $6$
6) $A•B•overlineC•overlineC+A•B•overlineC•overlineA$
Then one line 7: you claim as $A+ A = A$ that $A•B•overlineC•overlineC+A•B•overlineC•overlineA= A•B•overlineC•overlineA$
But $A•B•overlineC•overlineC$ and $A•B•overlineC•overlineA$ are not the same.
Instead use $Acdot A = A$ to get that $overlineC•overlineC = overlineC$ to get:
7) $A•B•overlineC+A•B•overlineC•overlineA$
Then distribute:
8) $A•B•overlineC(1 + overline A)$
9) $1 + A = 1 : A•B•overlineCcdot 1$
10) $1cdot A = A: Acdot Bcdot overlineC$.
...or....
8) $Aoverline A = 0: Acdot Bcdot overlineC + 0$
9) $A + 0 = A: Acdot Bcdot overlineC$
answered Aug 2 at 16:45
fleablood
60.1k22575
You have line $6$
6) $A•B•overlineC•overlineC+A•B•overlineC•overlineA$
Then one line 7: you claim as $A+ A = A$ that $A•B•overlineC•overlineC+A•B•overlineC•overlineA= A•B•overlineC•overlineA$
But $A•B•overlineC•overlineC$ and $A•B•overlineC•overlineA$ are not the same.
Instead use $Acdot A = A$ to get that $overlineC•overlineC = overlineC$ to get:
7) $A•B•overlineC+A•B•overlineC•overlineA$
Then distribute:
8) $A•B•overlineC(1 + overline A)$
9) $1 + A = 1 : A•B•overlineCcdot 1$
10) $1cdot A = A: Acdot Bcdot overlineC$.
...or....
8) $Aoverline A = 0: Acdot Bcdot overlineC + 0$
9) $A + 0 = A: Acdot Bcdot overlineC$
answered Aug 2 at 16:45
fleablood
60.1k22575
answered Aug 2 at 16:45
fleablood
60.1k22575
answered Aug 2 at 16:45
fleablood
60.1k22575
answered Aug 2 at 16:45
fleablood
60.1k22575
60.1k22575
add a comment |Â
add a comment |Â
up vote
0
down vote
Analogous with sets:
- $Acap((Bcap C)cup(Ccap A))^complementcap B$
- $Acap((Bcap C)^complementcap(Ccap A)^complement)cap B$
- $Acap((B^complementcup C^complement)cap(C^complementcup A^complement))cap B$
- $Acap((B^complementcap C^complement)cup(B^complementcap A^complement)cup (C^complementcap C^complement)cup(C^complementcap A^complement))cap B$
- $(Acap B^complementcap C^complementcap B)cup(Acap B^complementcap A^complementcap B)cup(Acap C^complementcap C^complementcap B)cup(Acap C^complementcap A^complementcap B)$
- $varnothingcupvarnothingcup(Acap C^complementcap C^complementcap B)cupvarnothing$
- $Acap C^complementcap B$
answered Aug 2 at 17:11
drhab
85.8k540118
add a comment |Â
up vote
0
down vote
Analogous with sets:
- $Acap((Bcap C)cup(Ccap A))^complementcap B$
- $Acap((Bcap C)^complementcap(Ccap A)^complement)cap B$
- $Acap((B^complementcup C^complement)cap(C^complementcup A^complement))cap B$
- $Acap((B^complementcap C^complement)cup(B^complementcap A^complement)cup (C^complementcap C^complement)cup(C^complementcap A^complement))cap B$
- $(Acap B^complementcap C^complementcap B)cup(Acap B^complementcap A^complementcap B)cup(Acap C^complementcap C^complementcap B)cup(Acap C^complementcap A^complementcap B)$
- $varnothingcupvarnothingcup(Acap C^complementcap C^complementcap B)cupvarnothing$
- $Acap C^complementcap B$
answered Aug 2 at 17:11
drhab
85.8k540118
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Analogous with sets:
- $Acap((Bcap C)cup(Ccap A))^complementcap B$
- $Acap((Bcap C)^complementcap(Ccap A)^complement)cap B$
- $Acap((B^complementcup C^complement)cap(C^complementcup A^complement))cap B$
- $Acap((B^complementcap C^complement)cup(B^complementcap A^complement)cup (C^complementcap C^complement)cup(C^complementcap A^complement))cap B$
- $(Acap B^complementcap C^complementcap B)cup(Acap B^complementcap A^complementcap B)cup(Acap C^complementcap C^complementcap B)cup(Acap C^complementcap A^complementcap B)$
- $varnothingcupvarnothingcup(Acap C^complementcap C^complementcap B)cupvarnothing$
- $Acap C^complementcap B$
answered Aug 2 at 17:11
drhab
85.8k540118
Analogous with sets:
- $Acap((Bcap C)cup(Ccap A))^complementcap B$
- $Acap((Bcap C)^complementcap(Ccap A)^complement)cap B$
- $Acap((B^complementcup C^complement)cap(C^complementcup A^complement))cap B$
- $Acap((B^complementcap C^complement)cup(B^complementcap A^complement)cup (C^complementcap C^complement)cup(C^complementcap A^complement))cap B$
- $(Acap B^complementcap C^complementcap B)cup(Acap B^complementcap A^complementcap B)cup(Acap C^complementcap C^complementcap B)cup(Acap C^complementcap A^complementcap B)$
- $varnothingcupvarnothingcup(Acap C^complementcap C^complementcap B)cupvarnothing$
- $Acap C^complementcap B$
answered Aug 2 at 17:11
drhab
85.8k540118
answered Aug 2 at 17:11
drhab
85.8k540118
answered Aug 2 at 17:11
drhab
85.8k540118
answered Aug 2 at 17:11
drhab
85.8k540118
85.8k540118
add a comment |Â
add a comment |Â
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Here a MathJax Tutorial
– saulspatz
Aug 2 at 16:17
"A+A=A : A•B•C•A." But you don't have "A•B•C•A+A•B•C•A" you have "A•B•C•C+A•B•C•A " and A•B•C•C$ne$A•B•C•A.
– fleablood
Aug 2 at 16:23
Thank you. I don't know why I thought that.
– John Smith
Aug 2 at 16:25
I edited and provided the proper Mathjax formula for your first expression ... can you edit and do the rest?
– Bram28
Aug 2 at 16:38
"I don't know why I thought that" A quasi-dyslexic proof reading over-sight. Frustratingly easy to make.
– fleablood
Aug 2 at 16:47