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what is difference between S1 and S2
S1:Any language L over an alphabet Σ,L^+=L-{ϵ} is always TRUE.

S2: Any language L over an alphabet Σ,L+=L^*-{ϵ} is always TRUE.

S3:The language L = {$a^nb^k$ : | n – k | = 2} is not regular
Which of the following statements is/are correct?
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S1:Any language L over an alphabet Σ,L^+=L-{ϵ} is always TRUE.<- This statement is false as we are are taken L^+ on one side which means that the language L can be concatenated one or more time but not zero times. but on the Leeft hand side we are just taking the language once and removing epsilon from it .So this is false.

S2: Any language L over an alphabet Σ,L+=L^*-{ϵ} is always TRUE.<- This statement is true.
L^* means the kleene closure of a language i.e the language 0 or more times . In simpler words:
L^*={ϵ,L^1,L^2,L^3,L^4........}
L^+={L^1,L^2,L^3,L^4........}
So lhs=rhs i.e:Σ,L+=L^*-{ϵ}
Note: In statement 1 on the rhs we have just taken the language and in statement 2 on the rhs we have taken the kleene closure of the language . This is the difference between the two.

4 Comments

@himgta
It is given that L is defined over an alphabet so I think L is not epsilon in this case so it isnt false.
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@ why L can not be epsilon?

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