CFL or not TOC

Question

Is L={$a^{n}b^{n}c^{m}$ : n>m} context free...???

Comments

why ? 

a^n b^n c^m --> n(a) = n(b) .......[only 1 comparison] 

a^ b^n c^m : n>m --> n(a) = n(b) and n(a) > n(c) --> here 2 comparisons at a time ---> so how CFL ?

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Okk got it.....thanks....

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if some one is thinking of pushing 2 disk in stack for a poping 1 disk for symbol b and the again poping disk for c !! pls think of comparisons----- n>m for a and c but what about |a|=|b|.

Answer 1

L={aⁿbⁿc^m : n>m};

After comparing a and b the stack will become empty, Now we don't know what is n, so we can't compare again with m.

Hence It is not CFL

Answer 2

L

anguage L={a^nb^nc^m: n>m} will not be CFL or DCFL.

Bcz when we are going to push all 'a' and going to pop all 'b' against 'a' then stack will contain only epsilon and there is no elements to compare with c inside stack as we are going to compare n to m.

That's why it will not be CFL.