No Ashutosh, L is regular. WE can understand it as: - If R is a non-regular part of regular language L, then the other part of L ( After removing R from L : ' L - R' ) must also be non regular such that when R comes together with it, both these parts cancel out each other's non-regularity. eg. L1={a^{m} b^{n}, m=n} is another way of writing R (L1=R). Now let L2 ={a^{m} b^{n }| m not equal to n }. Both L1 and L2 are non regular as both have infinite values of m,n and FA cant remember them. Now add L1 and L2. We get L1+L2={a^{m} b^{n}} i.e. no condition attached to m, n : all possible strings {a^m b^n} are accepted, which gives a regular language ,as L1 and L2 cancel each others non regularity when they come together.

**But I think we can definitely say that if L is regular and R is a non regular subset of L then L- R is also a non regular subset of L. **