How to prove a L=L(G) without knowing the L?
In cases like this, where you have a grammar and want to guess what the language is, a good way is to approach the problem systematically, by trying to derive some short strings. For example, if you had a grammar $G$ like this,
$$\begin{align}
S &amp;\rightarrow SA \tag{1}\\
S &amp;\rightarrow \epsilon \tag{2}\\
A &amp;\rightarrow aS \tag{3}\\
A &amp;\rightarrow bA \tag{4}\\
A &amp;\rightarrow \epsilon \tag{5}
\end{align}$$
then here are a few derivations, where the relevant production is labeled:
$$\begin{align}
\epsilon&amp;: S \stackrel{2}{\Longrightarrow}\epsilon\\
a&amp;: S \stackrel{1}{\Longrightarrow} SA\stackrel{2}{\Longrightarrow}A\stackrel{3}{\Longrightarrow}aS\stackrel{1}{\Longrightarrow}a\\
b&amp;: S \stackrel{1}{\Longrightarrow} SA\stackrel{2}{\Longrightarrow}A\stackrel{3}{\Longrightarrow}bA\stackrel{4}{\Longrightarrow}b\\
aa&amp;: S \stackrel{1}{\Longrightarrow} SA\stackrel{1}{\Longrightarrow}SAA\stackrel{2}{\Longrightarrow}AA\stackrel{3}{\Longrightarrow}aSA\stackrel{1}{\Longrightarrow}aA\stackrel{3}{\Longrightarrow}aaS\stackrel{2}{\Longrightarrow}aa\\
\end{align}$$
If you try this for, say $ab,ba,bb$, you'll find that they're all in $L$ which prompts a guess that this grammar generates all strings over the alphabet $\{a,b\}$. This isn't too hard to show, by induction on the length of a string in the language.

The proof goes like this:

Base. Show that the length-0 string, $\epsilon\in L(G)$. (We've done this already.)

Induction. Let $w$ be any string of length $n&gt;0$. Our inductive hypothesis is that any shorter string, $z$ is in $L(G)$, i.e., there is a derivation $S\stackrel{*}{\Longrightarrow}z$. Now either $w=za$ or $w=zb$. We'll do one case and leave the rest to you:
$$
S\stackrel{1}{\Longrightarrow}SA\stackrel{3}{\Longrightarrow}SaS\stackrel{2}{\Longrightarrow}Sa\stackrel{*}{\Longrightarrow}za = w
$$
so $w\in L(G)$, as required.