how do I simplify this particular boolean expression?
so I have spent nearly 5 hours trying to simplify this particular expression but I keep going round and round in circles. I have my hard copy notes to show you where I scribbled for hours and hours to no end. so please can someone please show me how can this be done?
$$
(bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})(bar{B}+D)(A+E)
$$
boolean-algebra
edited Nov 21 at 7:07
Matti P.
1,738413
asked Nov 20 at 11:28
ali farhad
32
add a comment |
so I have spent nearly 5 hours trying to simplify this particular expression but I keep going round and round in circles. I have my hard copy notes to show you where I scribbled for hours and hours to no end. so please can someone please show me how can this be done?
$$
(bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})(bar{B}+D)(A+E)
$$
boolean-algebra
edited Nov 21 at 7:07
Matti P.
1,738413
asked Nov 20 at 11:28
ali farhad
32
I would start opening the parentheses and then apply the known rules of booleans ... What is the simplest form that you have reached?
– Matti P.
Nov 20 at 11:35
i tried every single combination, even going as far as opening all the parentheses. it only added to the mess and I lost track of everything. I can't go more than 2 steps before I realize that I'm only undoing my earlier steps and somehow I find myself exactly where I'd started..
– ali farhad
Nov 20 at 11:39
Just use the law of distributivity. Its not a mess.
– Wuestenfux
Nov 20 at 11:43
did not work or maybe I can't work it out. I'm not too sure @Wuestenfux
– ali farhad
Nov 20 at 13:22
@MattiP. the simplest form i have been able to get is in this pic. I'm sorry I don't know how to type in LATEX and please excuse the bad writing and cutting. I'm kind of on edge with this particular problem. photos.app.goo.gl/L2Epwh2U1S67qBL68
– ali farhad
Nov 20 at 19:41
add a comment |
so I have spent nearly 5 hours trying to simplify this particular expression but I keep going round and round in circles. I have my hard copy notes to show you where I scribbled for hours and hours to no end. so please can someone please show me how can this be done?
$$
(bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})(bar{B}+D)(A+E)
$$
boolean-algebra
edited Nov 21 at 7:07
Matti P.
1,738413
asked Nov 20 at 11:28
ali farhad
32
so I have spent nearly 5 hours trying to simplify this particular expression but I keep going round and round in circles. I have my hard copy notes to show you where I scribbled for hours and hours to no end. so please can someone please show me how can this be done?
$$
(bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})(bar{B}+D)(A+E)
$$
boolean-algebra
boolean-algebra
edited Nov 21 at 7:07
Matti P.
1,738413
asked Nov 20 at 11:28
ali farhad
32
edited Nov 21 at 7:07
Matti P.
1,738413
asked Nov 20 at 11:28
ali farhad
32
edited Nov 21 at 7:07
Matti P.
1,738413
edited Nov 21 at 7:07
Matti P.
1,738413
edited Nov 21 at 7:07
Matti P.
1,738413
1,738413
asked Nov 20 at 11:28
ali farhad
32
asked Nov 20 at 11:28
ali farhad
32
asked Nov 20 at 11:28
ali farhad
32
32
I would start opening the parentheses and then apply the known rules of booleans ... What is the simplest form that you have reached?
– Matti P.
Nov 20 at 11:35
i tried every single combination, even going as far as opening all the parentheses. it only added to the mess and I lost track of everything. I can't go more than 2 steps before I realize that I'm only undoing my earlier steps and somehow I find myself exactly where I'd started..
– ali farhad
Nov 20 at 11:39
Just use the law of distributivity. Its not a mess.
– Wuestenfux
Nov 20 at 11:43
did not work or maybe I can't work it out. I'm not too sure @Wuestenfux
– ali farhad
Nov 20 at 13:22
@MattiP. the simplest form i have been able to get is in this pic. I'm sorry I don't know how to type in LATEX and please excuse the bad writing and cutting. I'm kind of on edge with this particular problem. photos.app.goo.gl/L2Epwh2U1S67qBL68
– ali farhad
Nov 20 at 19:41
add a comment |
I would start opening the parentheses and then apply the known rules of booleans ... What is the simplest form that you have reached?
– Matti P.
Nov 20 at 11:35
i tried every single combination, even going as far as opening all the parentheses. it only added to the mess and I lost track of everything. I can't go more than 2 steps before I realize that I'm only undoing my earlier steps and somehow I find myself exactly where I'd started..
– ali farhad
Nov 20 at 11:39
Just use the law of distributivity. Its not a mess.
– Wuestenfux
Nov 20 at 11:43
did not work or maybe I can't work it out. I'm not too sure @Wuestenfux
– ali farhad
Nov 20 at 13:22
@MattiP. the simplest form i have been able to get is in this pic. I'm sorry I don't know how to type in LATEX and please excuse the bad writing and cutting. I'm kind of on edge with this particular problem. photos.app.goo.gl/L2Epwh2U1S67qBL68
– ali farhad
Nov 20 at 19:41
I would start opening the parentheses and then apply the known rules of booleans ... What is the simplest form that you have reached?
– Matti P.
Nov 20 at 11:35
I would start opening the parentheses and then apply the known rules of booleans ... What is the simplest form that you have reached?
– Matti P.
Nov 20 at 11:35
i tried every single combination, even going as far as opening all the parentheses. it only added to the mess and I lost track of everything. I can't go more than 2 steps before I realize that I'm only undoing my earlier steps and somehow I find myself exactly where I'd started..
– ali farhad
Nov 20 at 11:39
i tried every single combination, even going as far as opening all the parentheses. it only added to the mess and I lost track of everything. I can't go more than 2 steps before I realize that I'm only undoing my earlier steps and somehow I find myself exactly where I'd started..
– ali farhad
Nov 20 at 11:39
Just use the law of distributivity. Its not a mess.
– Wuestenfux
Nov 20 at 11:43
Just use the law of distributivity. Its not a mess.
– Wuestenfux
Nov 20 at 11:43
did not work or maybe I can't work it out. I'm not too sure @Wuestenfux
– ali farhad
Nov 20 at 13:22
did not work or maybe I can't work it out. I'm not too sure @Wuestenfux
– ali farhad
Nov 20 at 13:22
@MattiP. the simplest form i have been able to get is in this pic. I'm sorry I don't know how to type in LATEX and please excuse the bad writing and cutting. I'm kind of on edge with this particular problem. photos.app.goo.gl/L2Epwh2U1S67qBL68
– ali farhad
Nov 20 at 19:41
@MattiP. the simplest form i have been able to get is in this pic. I'm sorry I don't know how to type in LATEX and please excuse the bad writing and cutting. I'm kind of on edge with this particular problem. photos.app.goo.gl/L2Epwh2U1S67qBL68
– ali farhad
Nov 20 at 19:41
add a comment |
1 Answer
1
active
oldest
votes
$$
begin{split}
& (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})(bar{B}+D)(A+E) \
= & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})(bar{B}+D)A \
+& (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})(bar{B}+D)E \
= & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})Abar{B} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})AD \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})bar{B}E \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})DE\
% ---------------
= & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D)Abar{B} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D)AD \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D)bar{B}E \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D)DE\
= & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)Abar{B}bar{E} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)AD bar{E}\
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)bar{B}underbrace{Ebar{E}}_{=0} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)Dunderbrace{Ebar{E}}_{=0}\
= & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)Abar{B}bar{E} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)AD bar{E}\
end{split}
$$
Now it's a good idea to take a littke break and comment on what I have done. I have approached the problem by opening carefully the parentheses and trying to match up variables with their negations (their product is zero). Following this logic and looking at the present expression, we see that the $bar{A}$ and $bar{E}$ in the beginning are also zero, because the expression is multiplied by $A E$ at the end. So now we have
$$
begin{split}
& (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)Abar{B}bar{E} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)AD bar{E}\
=& bar{B}(bar{A}+bar{C}+D)Abar{B}bar{E} \
+ & bar{B}(bar{A}+bar{C}+D)AD bar{E}\
=& (bar{C}+D)Abar{B}bar{E} \
+ & bar{B}(bar{C}+D)AD bar{E}\
=& Abar{B}left{ (bar{C}+D)bar{E}
+ (bar{C}+D)D bar{E}right} \
=& Abar{B}bar{E}(bar{C}+D)
end{split}
$$
You might want to recheck the details on this one.
answered Nov 21 at 7:12
Matti P.
1,738413
add a comment |
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1 Answer
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1 Answer
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$$
begin{split}
& (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})(bar{B}+D)(A+E) \
= & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})(bar{B}+D)A \
+& (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})(bar{B}+D)E \
= & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})Abar{B} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})AD \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})bar{B}E \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})DE\
% ---------------
= & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D)Abar{B} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D)AD \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D)bar{B}E \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D)DE\
= & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)Abar{B}bar{E} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)AD bar{E}\
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)bar{B}underbrace{Ebar{E}}_{=0} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)Dunderbrace{Ebar{E}}_{=0}\
= & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)Abar{B}bar{E} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)AD bar{E}\
end{split}
$$
Now it's a good idea to take a littke break and comment on what I have done. I have approached the problem by opening carefully the parentheses and trying to match up variables with their negations (their product is zero). Following this logic and looking at the present expression, we see that the $bar{A}$ and $bar{E}$ in the beginning are also zero, because the expression is multiplied by $A E$ at the end. So now we have
$$
begin{split}
& (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)Abar{B}bar{E} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)AD bar{E}\
=& bar{B}(bar{A}+bar{C}+D)Abar{B}bar{E} \
+ & bar{B}(bar{A}+bar{C}+D)AD bar{E}\
=& (bar{C}+D)Abar{B}bar{E} \
+ & bar{B}(bar{C}+D)AD bar{E}\
=& Abar{B}left{ (bar{C}+D)bar{E}
+ (bar{C}+D)D bar{E}right} \
=& Abar{B}bar{E}(bar{C}+D)
end{split}
$$
You might want to recheck the details on this one.
answered Nov 21 at 7:12
Matti P.
1,738413
add a comment |
$$
begin{split}
& (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})(bar{B}+D)(A+E) \
= & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})(bar{B}+D)A \
+& (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})(bar{B}+D)E \
= & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})Abar{B} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})AD \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})bar{B}E \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})DE\
% ---------------
= & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D)Abar{B} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D)AD \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D)bar{B}E \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D)DE\
= & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)Abar{B}bar{E} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)AD bar{E}\
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)bar{B}underbrace{Ebar{E}}_{=0} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)Dunderbrace{Ebar{E}}_{=0}\
= & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)Abar{B}bar{E} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)AD bar{E}\
end{split}
$$
Now it's a good idea to take a littke break and comment on what I have done. I have approached the problem by opening carefully the parentheses and trying to match up variables with their negations (their product is zero). Following this logic and looking at the present expression, we see that the $bar{A}$ and $bar{E}$ in the beginning are also zero, because the expression is multiplied by $A E$ at the end. So now we have
$$
begin{split}
& (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)Abar{B}bar{E} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)AD bar{E}\
=& bar{B}(bar{A}+bar{C}+D)Abar{B}bar{E} \
+ & bar{B}(bar{A}+bar{C}+D)AD bar{E}\
=& (bar{C}+D)Abar{B}bar{E} \
+ & bar{B}(bar{C}+D)AD bar{E}\
=& Abar{B}left{ (bar{C}+D)bar{E}
+ (bar{C}+D)D bar{E}right} \
=& Abar{B}bar{E}(bar{C}+D)
end{split}
$$
You might want to recheck the details on this one.
answered Nov 21 at 7:12
Matti P.
1,738413
add a comment |
$$
begin{split}
& (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})(bar{B}+D)(A+E) \
= & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})(bar{B}+D)A \
+& (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})(bar{B}+D)E \
= & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})Abar{B} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})AD \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})bar{B}E \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})DE\
% ---------------
= & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D)Abar{B} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D)AD \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D)bar{B}E \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D)DE\
= & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)Abar{B}bar{E} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)AD bar{E}\
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)bar{B}underbrace{Ebar{E}}_{=0} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)Dunderbrace{Ebar{E}}_{=0}\
= & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)Abar{B}bar{E} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)AD bar{E}\
end{split}
$$
Now it's a good idea to take a littke break and comment on what I have done. I have approached the problem by opening carefully the parentheses and trying to match up variables with their negations (their product is zero). Following this logic and looking at the present expression, we see that the $bar{A}$ and $bar{E}$ in the beginning are also zero, because the expression is multiplied by $A E$ at the end. So now we have
$$
begin{split}
& (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)Abar{B}bar{E} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)AD bar{E}\
=& bar{B}(bar{A}+bar{C}+D)Abar{B}bar{E} \
+ & bar{B}(bar{A}+bar{C}+D)AD bar{E}\
=& (bar{C}+D)Abar{B}bar{E} \
+ & bar{B}(bar{C}+D)AD bar{E}\
=& Abar{B}left{ (bar{C}+D)bar{E}
+ (bar{C}+D)D bar{E}right} \
=& Abar{B}bar{E}(bar{C}+D)
end{split}
$$
You might want to recheck the details on this one.
answered Nov 21 at 7:12
Matti P.
1,738413
$$
begin{split}
& (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})(bar{B}+D)(A+E) \
= & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})(bar{B}+D)A \
+& (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})(bar{B}+D)E \
= & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})Abar{B} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})AD \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})bar{B}E \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})DE\
% ---------------
= & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D)Abar{B} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D)AD \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D)bar{B}E \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D)DE\
= & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)Abar{B}bar{E} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)AD bar{E}\
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)bar{B}underbrace{Ebar{E}}_{=0} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)Dunderbrace{Ebar{E}}_{=0}\
= & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)Abar{B}bar{E} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)AD bar{E}\
end{split}
$$
Now it's a good idea to take a littke break and comment on what I have done. I have approached the problem by opening carefully the parentheses and trying to match up variables with their negations (their product is zero). Following this logic and looking at the present expression, we see that the $bar{A}$ and $bar{E}$ in the beginning are also zero, because the expression is multiplied by $A E$ at the end. So now we have
$$
begin{split}
& (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)Abar{B}bar{E} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)AD bar{E}\
=& bar{B}(bar{A}+bar{C}+D)Abar{B}bar{E} \
+ & bar{B}(bar{A}+bar{C}+D)AD bar{E}\
=& (bar{C}+D)Abar{B}bar{E} \
+ & bar{B}(bar{C}+D)AD bar{E}\
=& Abar{B}left{ (bar{C}+D)bar{E}
+ (bar{C}+D)D bar{E}right} \
=& Abar{B}bar{E}(bar{C}+D)
end{split}
$$
You might want to recheck the details on this one.
answered Nov 21 at 7:12
Matti P.
1,738413
answered Nov 21 at 7:12
Matti P.
1,738413
answered Nov 21 at 7:12
Matti P.
1,738413
answered Nov 21 at 7:12
Matti P.
1,738413
1,738413
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I would start opening the parentheses and then apply the known rules of booleans ... What is the simplest form that you have reached?
– Matti P.
Nov 20 at 11:35
i tried every single combination, even going as far as opening all the parentheses. it only added to the mess and I lost track of everything. I can't go more than 2 steps before I realize that I'm only undoing my earlier steps and somehow I find myself exactly where I'd started..
– ali farhad
Nov 20 at 11:39
Just use the law of distributivity. Its not a mess.
– Wuestenfux
Nov 20 at 11:43
did not work or maybe I can't work it out. I'm not too sure @Wuestenfux
– ali farhad
Nov 20 at 13:22
@MattiP. the simplest form i have been able to get is in this pic. I'm sorry I don't know how to type in LATEX and please excuse the bad writing and cutting. I'm kind of on edge with this particular problem. photos.app.goo.gl/L2Epwh2U1S67qBL68
– ali farhad
Nov 20 at 19:41