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Context free language is a grammar where its language or string that formed by context free grammar is supports pushdown deterministic automata.

This grammar is context free grammar is solved or proved by pumping lemma.

Here the pumping lemma is used to find the language is context free or not.

$\begin{array}{l}\left|\mathrm{z}\right|鈮�\mathrm{n}\\ \mathrm{z}=\mathrm{uvwxy}\\ \left|\mathrm{v}\right|鈮�1\end{array}$

Here at least 1 is not null.

$\begin{array}{l}\mathrm{L}=\left\{{\mathrm{a}}^{\mathrm{n}}{\mathrm{b}}^{\mathrm{n}}{\mathrm{c}}^{\mathrm{n}}\right|\mathrm{n}鈮�0\}\\ \mathrm{z}=\mathrm{uvwxy},\\ \mathrm{z}={\mathrm{a}}^{\mathrm{n}}{\mathrm{b}}^{\mathrm{n}}{\mathrm{c}}^{\mathrm{n}}\{\mathrm{n}=1,\mathrm{z}=\mathrm{abc}\}\\ \left|\mathrm{z}\right|=3\mathrm{n}>\mathrm{n}\end{array}$

If we equate both z then we get,

$\begin{array}{l}\mathrm{uvwxy}={\mathrm{a}}^{\mathrm{n}}{\mathrm{b}}^{\mathrm{n}}{\mathrm{c}}^{\mathrm{n}}\\ \left|\mathrm{vx}\right|鈮�1,\{\mathrm{n}鈮�|\mathrm{vx}|鈮�1\}\end{array}$

Then after that,

Let v or x is a part of -

$\begin{array}{l}\left({\mathrm{a}}^{\mathrm{i}}{\mathrm{b}}^{\mathrm{j}}\right)\mathrm{or}\left({\mathrm{b}}^{\mathrm{i}}{\mathrm{c}}^{\mathrm{j}}\right)\\ \mathrm{v}=\left({\mathrm{a}}^{\mathrm{i}}{\mathrm{b}}^{\mathrm{j}}\right)\\ {\mathrm{v}}^2=\left({\mathrm{a}}^{\mathrm{i}}{\mathrm{b}}^{\mathrm{j}}\right)\left({\mathrm{b}}^{\mathrm{i}}{\mathrm{c}}^{\mathrm{j}}\right)\\ {\mathrm{uv}}^2{\mathrm{wx}}^2\mathrm{y}={\mathrm{a}}^{\mathrm{m}}{\mathrm{b}}^{\mathrm{m}}{\mathrm{c}}^{{}^{\mathrm{m}}}\end{array}$

From this equation, it is proof that

${\mathrm{uv}}^2{\mathrm{wx}}^2\mathrm{y}鈭�\mathrm{L}$

Therefore, L is not context free language.

In the language given in the question if $C$is a context-free language and $R$is a regular language, then $\mathrm{C}鈭�\mathrm{R}$is context free. Therefore,$1*鈭�\text{鈥娾赌娾赌娾赌�}\mathrm{L}\left(\mathrm{G}\right)$, is context free.

The following pushdown automata decides the language of this problem.
鈥淥n input$鉄�G鉄�:$

1).Construct context-free language such that,

$\mathrm{L}\left(\mathrm{H}\right)=1*鈭�\text{鈥娾赌娾赌娾赌�}\mathrm{L}\left(\mathrm{G}\right).$

2). Test whether , $\mathrm{L}\left(\mathrm{H}\right)=\mathrm{桅}$

using the ${\mathrm{E}}_{\mathrm{CFG}}$ has a decider $\mathrm{R}$.

3). If$R$accepts, reject; if $R$rejects, accept.鈥�

Now, Let $\mathrm{PP}$ be an upper bound of the pumping lemma of$\mathrm{GG}$,

And here only need to check whether this string is defines as: $1^0,1^1,鈥�,{.....1}^{\mathrm{p}+\mathrm{p}\text{'}}鈭�\mathrm{G}$

If not, then certainly reject $\left\{\mathrm{G}\right\}\left\{\mathrm{G}\right\}.$

Otherwise, applying pumping lemma on $1^{\mathrm{k}}(\mathrm{p}鈮�\mathrm{k}鈮�\mathrm{p}+\mathrm{p}!),$

After that conclude that there exist substrings $\mathrm{u},\mathrm{v},\mathrm{w},\mathrm{x},\mathrm{y}$such that,

$1^{\mathrm{k}}=\mathrm{uvwxy}$, which means $\mathrm{u},\mathrm{v},\mathrm{w},\mathrm{x},\mathrm{y}$contain only $1\mathrm{s}$,

$\left|\mathrm{vwx}\right|鈮�\mathrm{pumping}\mathrm{length}鈮�\mathrm{p},$

$\left|\mathrm{vx}\right|鈮�1$

${\mathrm{uv}}^{\mathrm{n}}{\mathrm{wx}}^{\mathrm{n}}\mathrm{y}鈭�\mathrm{G},$

i.e.$1\left|\mathrm{uwy}\right|+\mathrm{n}\left|\mathrm{vx}\right|=1\mathrm{k}+(\mathrm{n}-1)\left|\mathrm{vx}\right|鈭�\mathrm{G}$ , for all $\mathrm{n}鈮�0$, which means$1\mathrm{k}+\mathrm{n}(\mathrm{p}!)鈭�\mathrm{G}1\mathrm{k}+\mathrm{n}(\mathrm{p}!)鈭�\mathrm{G}\mathrm{forall}\text{鈥夆赌�}\mathrm{n}鈮�0.$ hence, this language given in the question $\left\{\right\{\mathrm{G}\left\}\right|\mathrm{G}\mathrm{is}\mathrm{a}\mathrm{CFG}\mathrm{over}\{0,1\}\mathrm{and}1*鈯�\mathrm{L}\left(\mathrm{G}\right)\}$is decidable.