Proof a language is not regulary
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I need to proof that a language is not regulary, could this working?
$Proofquad thatquad Nquad =quad left{ { a }^{ m }{ a }^{ l }c{ b }^{ m+1 }|m,lquad in quad N right} quad withquad sum { =quad left{ a,b,c right} } is\ noquad regularquad languagequad withquad thequad PumpingLemma.\ \ Firstquad ofquad allquad iquad wouldquad likequad toquad write\ Nquad =quad left{ { a }^{ m }{ a }^{ l }c{ b }^{ m+1 }|m,lquad in quad N right} quad =quad left{ { a }^{ 2n }c{ b }^{ 2n }|nquad in quad N right} \ hopequad thatsquad ok?.\ \ Proof:\ Bequad nin Nquad arbitraryquad butquad firm.quad Wequad choosequad thequad wordquad wquad =quad { a }^{ 2n }c{ b }^{ 2n }\ withquad win quad Nquad andquad |w|ge n.\ Bequad w=xyzquad aquad arbitraryquad decompositionquad withquad yneq lambda quad andquad |xy|le n.\ Thenquad wequad havequad xquad =quad { a }^{ 2i },quad yquad =quad { a }^{ 2j }quad andquad zquad =quad { a }^{ 2n-2i-2j }c{ b }^{ 2n }quad forquad jneq 0quad andquad 2(i+j)le 2n.\ Wequad choosequad kquad =quad 0.quad Thenquad wequad havequad x{ y }^{ 0 }zquad ={ quad a }^{ 2n-2i }c{ b }^{ 2n }.quad =>quad x{ y }^{ 0 }zquad notin Nquad because\ 2n-2ineq 2nquad forquad jneq 0.quad \ =>quad Nquad isquad noquad regularyquad language.\ $
proof-verification regular-language
asked Jan 17 at 20:09
Lisa.NeustLisa.Neust
12
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add a comment |
$begingroup$
I need to proof that a language is not regulary, could this working?
$Proofquad thatquad Nquad =quad left{ { a }^{ m }{ a }^{ l }c{ b }^{ m+1 }|m,lquad in quad N right} quad withquad sum { =quad left{ a,b,c right} } is\ noquad regularquad languagequad withquad thequad PumpingLemma.\ \ Firstquad ofquad allquad iquad wouldquad likequad toquad write\ Nquad =quad left{ { a }^{ m }{ a }^{ l }c{ b }^{ m+1 }|m,lquad in quad N right} quad =quad left{ { a }^{ 2n }c{ b }^{ 2n }|nquad in quad N right} \ hopequad thatsquad ok?.\ \ Proof:\ Bequad nin Nquad arbitraryquad butquad firm.quad Wequad choosequad thequad wordquad wquad =quad { a }^{ 2n }c{ b }^{ 2n }\ withquad win quad Nquad andquad |w|ge n.\ Bequad w=xyzquad aquad arbitraryquad decompositionquad withquad yneq lambda quad andquad |xy|le n.\ Thenquad wequad havequad xquad =quad { a }^{ 2i },quad yquad =quad { a }^{ 2j }quad andquad zquad =quad { a }^{ 2n-2i-2j }c{ b }^{ 2n }quad forquad jneq 0quad andquad 2(i+j)le 2n.\ Wequad choosequad kquad =quad 0.quad Thenquad wequad havequad x{ y }^{ 0 }zquad ={ quad a }^{ 2n-2i }c{ b }^{ 2n }.quad =>quad x{ y }^{ 0 }zquad notin Nquad because\ 2n-2ineq 2nquad forquad jneq 0.quad \ =>quad Nquad isquad noquad regularyquad language.\ $
proof-verification regular-language
asked Jan 17 at 20:09
Lisa.NeustLisa.Neust
12
$endgroup$
add a comment |
$begingroup$
I need to proof that a language is not regulary, could this working?
$Proofquad thatquad Nquad =quad left{ { a }^{ m }{ a }^{ l }c{ b }^{ m+1 }|m,lquad in quad N right} quad withquad sum { =quad left{ a,b,c right} } is\ noquad regularquad languagequad withquad thequad PumpingLemma.\ \ Firstquad ofquad allquad iquad wouldquad likequad toquad write\ Nquad =quad left{ { a }^{ m }{ a }^{ l }c{ b }^{ m+1 }|m,lquad in quad N right} quad =quad left{ { a }^{ 2n }c{ b }^{ 2n }|nquad in quad N right} \ hopequad thatsquad ok?.\ \ Proof:\ Bequad nin Nquad arbitraryquad butquad firm.quad Wequad choosequad thequad wordquad wquad =quad { a }^{ 2n }c{ b }^{ 2n }\ withquad win quad Nquad andquad |w|ge n.\ Bequad w=xyzquad aquad arbitraryquad decompositionquad withquad yneq lambda quad andquad |xy|le n.\ Thenquad wequad havequad xquad =quad { a }^{ 2i },quad yquad =quad { a }^{ 2j }quad andquad zquad =quad { a }^{ 2n-2i-2j }c{ b }^{ 2n }quad forquad jneq 0quad andquad 2(i+j)le 2n.\ Wequad choosequad kquad =quad 0.quad Thenquad wequad havequad x{ y }^{ 0 }zquad ={ quad a }^{ 2n-2i }c{ b }^{ 2n }.quad =>quad x{ y }^{ 0 }zquad notin Nquad because\ 2n-2ineq 2nquad forquad jneq 0.quad \ =>quad Nquad isquad noquad regularyquad language.\ $
proof-verification regular-language
asked Jan 17 at 20:09
Lisa.NeustLisa.Neust
12
$endgroup$
I need to proof that a language is not regulary, could this working?
$Proofquad thatquad Nquad =quad left{ { a }^{ m }{ a }^{ l }c{ b }^{ m+1 }|m,lquad in quad N right} quad withquad sum { =quad left{ a,b,c right} } is\ noquad regularquad languagequad withquad thequad PumpingLemma.\ \ Firstquad ofquad allquad iquad wouldquad likequad toquad write\ Nquad =quad left{ { a }^{ m }{ a }^{ l }c{ b }^{ m+1 }|m,lquad in quad N right} quad =quad left{ { a }^{ 2n }c{ b }^{ 2n }|nquad in quad N right} \ hopequad thatsquad ok?.\ \ Proof:\ Bequad nin Nquad arbitraryquad butquad firm.quad Wequad choosequad thequad wordquad wquad =quad { a }^{ 2n }c{ b }^{ 2n }\ withquad win quad Nquad andquad |w|ge n.\ Bequad w=xyzquad aquad arbitraryquad decompositionquad withquad yneq lambda quad andquad |xy|le n.\ Thenquad wequad havequad xquad =quad { a }^{ 2i },quad yquad =quad { a }^{ 2j }quad andquad zquad =quad { a }^{ 2n-2i-2j }c{ b }^{ 2n }quad forquad jneq 0quad andquad 2(i+j)le 2n.\ Wequad choosequad kquad =quad 0.quad Thenquad wequad havequad x{ y }^{ 0 }zquad ={ quad a }^{ 2n-2i }c{ b }^{ 2n }.quad =>quad x{ y }^{ 0 }zquad notin Nquad because\ 2n-2ineq 2nquad forquad jneq 0.quad \ =>quad Nquad isquad noquad regularyquad language.\ $
proof-verification regular-language
proof-verification regular-language
asked Jan 17 at 20:09
Lisa.NeustLisa.Neust
12
asked Jan 17 at 20:09
Lisa.NeustLisa.Neust
12
edited Jan 17 at 22:31
Lisa.Neust
asked Jan 17 at 20:09
Lisa.NeustLisa.Neust
12
asked Jan 17 at 20:09
Lisa.NeustLisa.Neust
12
asked Jan 17 at 20:09
Lisa.NeustLisa.Neust
12
12
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