Toc Gate 2018 sample practice

Question

Consider 2 regular expression:
i. ϕ* + a+ + b+ + (a + b)+ → r1
ii. ϕ+ + a* + b* + (a + b)* → r2

(a) L(r1) = L(r2) (b) L(r1) ⊆ L(r2)
(c) L(r1) ⊇ L(r2) (d) None of above
Solution: Option (a)

 

how need explanation? wont answer be c?

Comments

option a) is correct.

L(R1) = L(R2) = (a+b)*

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(1) epsilon+a+b+(a+b)

(2) phai+epsilon+epsilon+epsilon minimum string genrated

Answer 1

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Comments

but in r1 + is given so dont = means 1 or more ? while * means 0 or more then wont + be subset of *?

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but in R1, there is also ϕ* = epsilon, now everthing else could be generated by (a+b)⁺