how do I simplify this particular boolean expression?
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0
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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 yesterday
Matti P.
1,625413
asked 2 days ago
ali farhad
32
New contributor
ali farhad is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
up vote
0
down vote
favorite
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 yesterday
Matti P.
1,625413
asked 2 days ago
ali farhad
32
New contributor
ali farhad is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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.
2 days ago
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
2 days ago
Just use the law of distributivity. Its not a mess.
– Wuestenfux
2 days ago
did not work or maybe I can't work it out. I'm not too sure @Wuestenfux
– ali farhad
2 days ago
@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
2 days ago
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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 yesterday
Matti P.
1,625413
asked 2 days ago
ali farhad
32
New contributor
ali farhad is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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 yesterday
Matti P.
1,625413
asked 2 days ago
ali farhad
32
New contributor
ali farhad is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Matti P.
1,625413
asked 2 days ago
ali farhad
32
New contributor
ali farhad is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Matti P.
1,625413
edited yesterday
Matti P.
1,625413
edited yesterday
Matti P.
1,625413
1,625413
asked 2 days ago
ali farhad
32
New contributor
ali farhad is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
ali farhad
32
asked 2 days ago
ali farhad
32
32
New contributor
ali farhad is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
ali farhad is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
ali farhad is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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.
2 days ago
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
2 days ago
Just use the law of distributivity. Its not a mess.
– Wuestenfux
2 days ago
did not work or maybe I can't work it out. I'm not too sure @Wuestenfux
– ali farhad
2 days ago
@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
2 days ago
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.
2 days ago
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
2 days ago
Just use the law of distributivity. Its not a mess.
– Wuestenfux
2 days ago
did not work or maybe I can't work it out. I'm not too sure @Wuestenfux
– ali farhad
2 days ago
@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
2 days ago
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.
2 days ago
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.
2 days ago
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
2 days ago
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
2 days ago
Just use the law of distributivity. Its not a mess.
– Wuestenfux
2 days ago
Just use the law of distributivity. Its not a mess.
– Wuestenfux
2 days ago
did not work or maybe I can't work it out. I'm not too sure @Wuestenfux
– ali farhad
2 days ago
did not work or maybe I can't work it out. I'm not too sure @Wuestenfux
– ali farhad
2 days ago
@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
2 days ago
@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
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
$$
begin{split}
& (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})(bar{B}+D)(A+E) \
= & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})(bar{B}+D)A \
+& (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})(bar{B}+D)E \
= & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})Abar{B} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})AD \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})bar{B}E \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})DE\
% ---------------
= & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D)Abar{B} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D)AD \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D)bar{B}E \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D)DE\
= & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)Abar{B}bar{E} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)AD bar{E}\
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)bar{B}underbrace{Ebar{E}}_{=0} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)Dunderbrace{Ebar{E}}_{=0}\
= & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)Abar{B}bar{E} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)AD bar{E}\
end{split}
$$
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 yesterday
Matti P.
1,625413
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
$$
begin{split}
& (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})(bar{B}+D)(A+E) \
= & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})(bar{B}+D)A \
+& (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})(bar{B}+D)E \
= & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})Abar{B} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})AD \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})bar{B}E \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})DE\
% ---------------
= & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D)Abar{B} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D)AD \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D)bar{B}E \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D)DE\
= & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)Abar{B}bar{E} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)AD bar{E}\
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)bar{B}underbrace{Ebar{E}}_{=0} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)Dunderbrace{Ebar{E}}_{=0}\
= & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)Abar{B}bar{E} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)AD bar{E}\
end{split}
$$
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 yesterday
Matti P.
1,625413
add a comment |
up vote
0
down vote
accepted
$$
begin{split}
& (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})(bar{B}+D)(A+E) \
= & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})(bar{B}+D)A \
+& (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})(bar{B}+D)E \
= & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})Abar{B} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})AD \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})bar{B}E \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})DE\
% ---------------
= & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D)Abar{B} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D)AD \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D)bar{B}E \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D)DE\
= & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)Abar{B}bar{E} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)AD bar{E}\
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)bar{B}underbrace{Ebar{E}}_{=0} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)Dunderbrace{Ebar{E}}_{=0}\
= & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)Abar{B}bar{E} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)AD bar{E}\
end{split}
$$
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 yesterday
Matti P.
1,625413
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
$$
begin{split}
& (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})(bar{B}+D)(A+E) \
= & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})(bar{B}+D)A \
+& (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})(bar{B}+D)E \
= & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})Abar{B} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})AD \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})bar{B}E \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})DE\
% ---------------
= & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D)Abar{B} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D)AD \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D)bar{B}E \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D)DE\
= & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)Abar{B}bar{E} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)AD bar{E}\
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)bar{B}underbrace{Ebar{E}}_{=0} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)Dunderbrace{Ebar{E}}_{=0}\
= & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)Abar{B}bar{E} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)AD bar{E}\
end{split}
$$
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 yesterday
Matti P.
1,625413
$$
begin{split}
& (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})(bar{B}+D)(A+E) \
= & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})(bar{B}+D)A \
+& (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})(bar{B}+D)E \
= & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})Abar{B} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})AD \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})bar{B}E \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D+bar{E})DE\
% ---------------
= & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D)Abar{B} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D)AD \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D)bar{B}E \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)(C+D)DE\
= & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)Abar{B}bar{E} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)AD bar{E}\
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)bar{B}underbrace{Ebar{E}}_{=0} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)Dunderbrace{Ebar{E}}_{=0}\
= & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)Abar{B}bar{E} \
+ & (bar{A} + bar{B} + E)(bar{A}+bar{C}+D)AD bar{E}\
end{split}
$$
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 yesterday
Matti P.
1,625413
answered yesterday
Matti P.
1,625413
answered yesterday
Matti P.
1,625413
answered yesterday
Matti P.
1,625413
1,625413
add a comment |
add a comment |
ali farhad is a new contributor. Be nice, and check out our Code of Conduct.
ali farhad is a new contributor. Be nice, and check out our Code of Conduct.
ali farhad is a new contributor. Be nice, and check out our Code of Conduct.
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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.
2 days ago
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
2 days ago
Just use the law of distributivity. Its not a mess.
– Wuestenfux
2 days ago
did not work or maybe I can't work it out. I'm not too sure @Wuestenfux
– ali farhad
2 days ago
@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
2 days ago