Proving 2 languages are equal.
0
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 Nov 19 '18 at 5:06
Alan Bui
163
|
show 6 more comments
0
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 Nov 19 '18 at 5:06
Alan Bui
163
Have you tried some simple example? If the alphabet is ${1}$, $A=11$ and $B=1$, what do you get?
– Fabio Somenzi
Nov 19 '18 at 5:32
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
Nov 19 '18 at 5:53
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
Nov 19 '18 at 5:56
@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
Nov 19 '18 at 6:00
@keoxkeox The alphabet is supposed to have only one letter; otherwise the claim is obviously wrong.
– Fabio Somenzi
Nov 19 '18 at 6:01
|
show 6 more comments
0
0
0
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 Nov 19 '18 at 5:06
Alan Bui
163
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 Nov 19 '18 at 5:06
Alan Bui
163
asked Nov 19 '18 at 5:06
Alan Bui
163
edited Nov 19 '18 at 5:55
asked Nov 19 '18 at 5:06
Alan Bui
163
asked Nov 19 '18 at 5:06
Alan Bui
163
asked Nov 19 '18 at 5:06
Alan Bui
163
163
Have you tried some simple example? If the alphabet is ${1}$, $A=11$ and $B=1$, what do you get?
– Fabio Somenzi
Nov 19 '18 at 5:32
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
Nov 19 '18 at 5:53
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
Nov 19 '18 at 5:56
@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
Nov 19 '18 at 6:00
@keoxkeox The alphabet is supposed to have only one letter; otherwise the claim is obviously wrong.
– Fabio Somenzi
Nov 19 '18 at 6:01
|
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
Nov 19 '18 at 5:32
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
Nov 19 '18 at 5:53
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
Nov 19 '18 at 5:56
@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
Nov 19 '18 at 6:00
@keoxkeox The alphabet is supposed to have only one letter; otherwise the claim is obviously wrong.
– Fabio Somenzi
Nov 19 '18 at 6:01
Have you tried some simple example? If the alphabet is ${1}$, $A=11$ and $B=1$, what do you get?
– Fabio Somenzi
Nov 19 '18 at 5:32
Have you tried some simple example? If the alphabet is ${1}$, $A=11$ and $B=1$, what do you get?
– Fabio Somenzi
Nov 19 '18 at 5:32
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
Nov 19 '18 at 5:53
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
Nov 19 '18 at 5:53
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
Nov 19 '18 at 5:56
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
Nov 19 '18 at 5:56
@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
Nov 19 '18 at 6:00
@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
Nov 19 '18 at 6:00
@keoxkeox The alphabet is supposed to have only one letter; otherwise the claim is obviously wrong.
– Fabio Somenzi
Nov 19 '18 at 6:01
@keoxkeox The alphabet is supposed to have only one letter; otherwise the claim is obviously wrong.
– Fabio Somenzi
Nov 19 '18 at 6:01
|
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
Nov 19 '18 at 5:32
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
Nov 19 '18 at 5:53
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
Nov 19 '18 at 5:56
@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
Nov 19 '18 at 6:00
@keoxkeox The alphabet is supposed to have only one letter; otherwise the claim is obviously wrong.
– Fabio Somenzi
Nov 19 '18 at 6:01