Mixing
Proving 2 languages are equal.
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I am quite new to Discrete Math and having lots of troubles solving the problems in it.
Currently, I am struggling with a problem:
Prove by induction that if $A$ and $B$ are regular expressions over one-letter alphabet and if $n$ is any natural, prove that languages $(AB)^n$ and $A^nB^n$ are equal.
Thanks
discrete-mathematics regular-expressions
asked yesterday
Alan Bui
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up vote
0
down vote
favorite
I am quite new to Discrete Math and having lots of troubles solving the problems in it.
Currently, I am struggling with a problem:
Prove by induction that if $A$ and $B$ are regular expressions over one-letter alphabet and if $n$ is any natural, prove that languages $(AB)^n$ and $A^nB^n$ are equal.
Thanks
discrete-mathematics regular-expressions
asked yesterday
Alan Bui
113
New contributor
Alan Bui is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Have you tried some simple example? If the alphabet is ${1}$, $A=11$ and $B=1$, what do you get?
– Fabio Somenzi
yesterday
Sorry to say this but I am pretty much lost and do not know where to start my thing. I need some examples so I can follow it.
– Alan Bui
yesterday
If $A$ and $B$ are as above and $n=2$, $(AB)^n = (111)^2 = 111111$ and $A^nB^n = (11)^2(1)^2 = 111111$.
– Fabio Somenzi
yesterday
@FabioSomenzi can you please do an example with $A=a$ and $B=b$ where $a neq b$ or $a$ and $b$ are not concats of each other?
– keoxkeox
23 hours ago
@keoxkeox The alphabet is supposed to have only one letter; otherwise the claim is obviously wrong.
– Fabio Somenzi
23 hours ago
|
show 6 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am quite new to Discrete Math and having lots of troubles solving the problems in it.
Currently, I am struggling with a problem:
Prove by induction that if $A$ and $B$ are regular expressions over one-letter alphabet and if $n$ is any natural, prove that languages $(AB)^n$ and $A^nB^n$ are equal.
Thanks
discrete-mathematics regular-expressions
asked yesterday
Alan Bui
113
New contributor
Alan Bui is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I am quite new to Discrete Math and having lots of troubles solving the problems in it.
Currently, I am struggling with a problem:
Prove by induction that if $A$ and $B$ are regular expressions over one-letter alphabet and if $n$ is any natural, prove that languages $(AB)^n$ and $A^nB^n$ are equal.
Thanks
discrete-mathematics regular-expressions
discrete-mathematics regular-expressions
asked yesterday
Alan Bui
113
New contributor
Alan Bui is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Alan Bui
113
New contributor
Alan Bui is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
asked yesterday
Alan Bui
113
New contributor
Alan Bui is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Alan Bui
113
asked yesterday
Alan Bui
113
113
New contributor
Alan Bui 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 Bui is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Alan Bui is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Have you tried some simple example? If the alphabet is ${1}$, $A=11$ and $B=1$, what do you get?
– Fabio Somenzi
yesterday
Sorry to say this but I am pretty much lost and do not know where to start my thing. I need some examples so I can follow it.
– Alan Bui
yesterday
If $A$ and $B$ are as above and $n=2$, $(AB)^n = (111)^2 = 111111$ and $A^nB^n = (11)^2(1)^2 = 111111$.
– Fabio Somenzi
yesterday
@FabioSomenzi can you please do an example with $A=a$ and $B=b$ where $a neq b$ or $a$ and $b$ are not concats of each other?
– keoxkeox
23 hours ago
@keoxkeox The alphabet is supposed to have only one letter; otherwise the claim is obviously wrong.
– Fabio Somenzi
23 hours ago
|
show 6 more comments
Have you tried some simple example? If the alphabet is ${1}$, $A=11$ and $B=1$, what do you get?
– Fabio Somenzi
yesterday
Sorry to say this but I am pretty much lost and do not know where to start my thing. I need some examples so I can follow it.
– Alan Bui
yesterday
If $A$ and $B$ are as above and $n=2$, $(AB)^n = (111)^2 = 111111$ and $A^nB^n = (11)^2(1)^2 = 111111$.
– Fabio Somenzi
yesterday
@FabioSomenzi can you please do an example with $A=a$ and $B=b$ where $a neq b$ or $a$ and $b$ are not concats of each other?
– keoxkeox
23 hours ago
@keoxkeox The alphabet is supposed to have only one letter; otherwise the claim is obviously wrong.
– Fabio Somenzi
23 hours ago
Have you tried some simple example? If the alphabet is ${1}$, $A=11$ and $B=1$, what do you get?
– Fabio Somenzi
yesterday
Have you tried some simple example? If the alphabet is ${1}$, $A=11$ and $B=1$, what do you get?
– Fabio Somenzi
yesterday
Sorry to say this but I am pretty much lost and do not know where to start my thing. I need some examples so I can follow it.
– Alan Bui
yesterday
Sorry to say this but I am pretty much lost and do not know where to start my thing. I need some examples so I can follow it.
– Alan Bui
yesterday
If $A$ and $B$ are as above and $n=2$, $(AB)^n = (111)^2 = 111111$ and $A^nB^n = (11)^2(1)^2 = 111111$.
– Fabio Somenzi
yesterday
If $A$ and $B$ are as above and $n=2$, $(AB)^n = (111)^2 = 111111$ and $A^nB^n = (11)^2(1)^2 = 111111$.
– Fabio Somenzi
yesterday
@FabioSomenzi can you please do an example with $A=a$ and $B=b$ where $a neq b$ or $a$ and $b$ are not concats of each other?
– keoxkeox
23 hours ago
@FabioSomenzi can you please do an example with $A=a$ and $B=b$ where $a neq b$ or $a$ and $b$ are not concats of each other?
– keoxkeox
23 hours ago
@keoxkeox The alphabet is supposed to have only one letter; otherwise the claim is obviously wrong.
– Fabio Somenzi
23 hours ago
@keoxkeox The alphabet is supposed to have only one letter; otherwise the claim is obviously wrong.
– Fabio Somenzi
23 hours ago
|
show 6 more comments
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Have you tried some simple example? If the alphabet is ${1}$, $A=11$ and $B=1$, what do you get?
– Fabio Somenzi
yesterday
Sorry to say this but I am pretty much lost and do not know where to start my thing. I need some examples so I can follow it.
– Alan Bui
yesterday
If $A$ and $B$ are as above and $n=2$, $(AB)^n = (111)^2 = 111111$ and $A^nB^n = (11)^2(1)^2 = 111111$.
– Fabio Somenzi
yesterday
@FabioSomenzi can you please do an example with $A=a$ and $B=b$ where $a neq b$ or $a$ and $b$ are not concats of each other?
– keoxkeox
23 hours ago
@keoxkeox The alphabet is supposed to have only one letter; otherwise the claim is obviously wrong.
– Fabio Somenzi
23 hours ago