a. case 1: when i=1 F=$ab^nc^n$. it is DCFG
case 2: when i$\neq$1 F=$a^ib^jc^k$ | i,j,k$\geq$0
this is aaa*b*c* $\bigcup$ b*c*
F=$ab^nc^n$ $\bigcup$ {aaa*b*c* $\bigcup$ b*c*}
=DCFG $\bigcup$ regular
=DCFG
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b. Three conditions of Pumping Lemma
for any regular language L, there exists an integer p(pumping length) such that for all strings w$\varepsilon$L and |w|$\geq$p (i.e.
the length of string is atleast p), there exists x, y, z such that w=xyz (the string can be broken down into three parts) and
1. |xy|$\leq$p length of xy must not exceed the pumping length
2. |y|$\geq$1 y cannot be null string
3. for all integer i$\geq$0, $xy^iz\varepsilon L$
taking p=4
case 1: x=ε, y=bbbb, z=ε
w=$(bbbb)^i$
for i=0, w=ε
for i=2, w=bbbbbbbb
for all i, w belongs to L
It behaves like regular language
case 2: x=aa, y=bb, z=c
w=$aa(bb)^ic$
i=0, w=aac
i=5, w=aabbbbbbbbbbc
for all i, w belongs to L
Pumping Lemma test will fail when x=a, y=bbb, z=ccc