As far as I know, this is textbook stuff so the proofs are obviously not found in articles. However if I remember correctly, you get to Chomsky normal form by first replacing any terminals in something like
$$A \rightarrow BCst$$
by some extra nonterminal symbols
$$A \rightarrow BCST$$
$$S \rightarrow s$$
$$T \rightarrow t$$
and then in the second step you replace longer rules like
$$A\rightarrow BCD$$
with again additional nonterminals
$$A\rightarrow BA'$$
$$A'\rightarrow CD$$
I think it is easy to see that each replacement step only needs $O(n)$ of work per rule, so you end up with $O(n^2)$, since the number of rules is clearly bounded by $n$.
HINT: There are two slightly different definitions of Chomsky normal form. It appears that you’re using the one that simply requires each production to be of the form $X\to YZ$ or $X\to a$, rather than the one in Wikipedia; this causes no problems, since the lack of $\lambda$-productions means that $\lambda$ isn’t in the language anyway.
The algorithm given in that article for converting $G$ to $G'$ still works; we’re just in the fortunate situation of being able to ignore parts of it as not applicable. $G$ has no $\lambda$-productions or unit productions, and your definition of Chomsky normal form doesn’t require elimination of the initial symbol from the righthand side (the START step), so the last two steps, DEL and UNIT, don’t arise, and we need only perform the TERM and BIN steps.
- Show that the TERM step adds at most $|T|$ productions to the grammar.
After the TERM step has been completed, every production is of one of two kinds: $X\to a$ (where $X$ might be one of the new non-terminals introduced in the TERM step), or $X\to Y_1\ldots Y_n$, where $n\ge 2$. Productions of the form $X\to Y_1Y_2$ are fine, but we have to get rid of the ones with $n\ge 3$.
- Show that the procedure used in the BIN step replaces a production $X\to Y_1\ldots Y_n$ with $n-1$ productions of the form $Z\to UV$.
- Conclude that $G'$ has at most $|T|+(K-1)|P|$ productions.
If you were using the definition of Chomsky normal form given in the Wikipedia article, the grammar with initial symbol $S$, terminal symbols $a$ and $b$, and productions $S\to Sa\mid bb$ would be a counterexample to the theorem. Clearly $|P|=|T|=K=2$, so $(K-1)|P|+|T|=4$. However, the smallest equivalent grammar in (this version of) Chomsky normal form has six productions:
$$\begin{align*}
&S\to BB\mid XA\\
&X\to BB\mid XA\\
&A\to a\\
&B\to b
\end{align*}$$
If we allow $S$ to appear on the righthand side, we can reduce this to four, as in the theorem:
$$\begin{align*}
&S\to BB\mid SA\\
&A\to a\\
&B\to b
\end{align*}$$
Best Answer
Conversion from Context Free Grammar to Chomsky Normal Form : (I ll tell you the steps and will also solve the example you asked simultaneously)
Step 1 : Introduce New Non-terminal $S_0$ and make it derive the start variable which is S
Thus
Step 2 : Eliminate all $\varepsilon$ transitions
THus we need to eliminate B -> $\varepsilon$
For this we must to replace B with $\varepsilon$ in RHS of every production having B
THus we get,
Now new $\varepsilon$ transition is introduced which is A -> $\varepsilon$ .. thus we need to eliminate it too
Step 3 : Eliminate all Unit transitions i.e. those productions having exactly one non-terminal in RHS .
thus we need to eliminate A -> B , A->S ,$S_0$ -> S
THus,first removing A-> B
NOw removing A-> S
NOw removing $S_0$ -> S
Step 4 : Now eliminate all the productions that are non in CNF
The above CFG is in CNF .
:)