TOC, Language Identification

Question

Comments

the answer is DCFL.because L1-L2 will contain al string which has atleast one b so L1-L2=(a^nb^+c^n) which is DCFL

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So the ans is ?

Answer 1

L1= {epsilon,ac,abc,aabcc,b,bb,bbb.....}

L2={epsilon, ab,aabb,aaabbb..............}

as we can see only epsilon is common

L1-L2 ={ ac,abc,aabcc,b,bb,bbb......}

={a^n b^m c^n | m>=0,n>0 or m>0,n>=0} which is DCFL

Comments

@akash

If we take L1-L2=L1 intersection L2'

So L2'={abb,ba,aab,b,a,bb,...} (anything expct L2)

L1={abc,abbc,b,bb,aabcc,...}

Now common between them

{b,bb,bbb...} wiz. b+

So wouldnt that be Regular ?

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@abbas2131

the above lang. also contain b* if we put n=0,m>0

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But its about intersection og L1 and L2'

So why re we taking any string with c in it ffom L1, because L2' wont have it.