I have to find out if the grammar $G=(\{A,B\},\{x,y\},A)$, where the productions of $P$ are given by $$\begin{array}{ccc}A&\rightarrow &x\:A/y\:B\\B&\rightarrow &x\:B/x\end{array}$$ is type 3 and generates a finite language.
When it says "type 3" it refers to the Chomsky Hierarchy of the types of grammars, and there are 4 types, where $G_3\subset G_2\subset G_1\subset G_0$. As I think it is in type 3, I am going to write the definition that I have:
A grammar is Type 3 if all production $x\to y$:
$x \text{ it is a single symbol of the non-terminal elements;}\\ y \text{ is }\left\{\begin{array}{l}\text{The concatenation of 2 symbols, one of them being terminal,}\\\qquad\qquad\text{or}\\\text{A single terminal symbol,}\\\qquad\qquad\text{or}\\\text{It is the null word.}\end{array}\right.$
With this I can affirm that both production $A\to x\: A/y\: B$ and production $B\to x\: B/x$ meet these requirements, and therefore the grammar is type 3.
With respect to what generates a finite language, I did not start making the derivation tree, but I understand that the production is being called "recursively"; that is, it will always continue to generate words (for example, $A$ has two outputs: $x\: A$ - it is generated on itself - and $y\: B$). Therefore, it will never generate a finite language.
Is it right?
Also, a very short question: If we have for example that a generated language is $L(G)=\left\{0^{\ast} 1^{\ast}\right\}$, is it an infinite language?
Thanks!
You are right, the grammar is of type 3. Your definition though seems flawed. If you allow the terminals to appear on both sides (though in different productions) of the non-terminal, then the type of grammar generates all linear languages, which are more than the regular ones.
Also your answer to part two of the question is correct. A detail your argumentation is missing is the fact that the unbounded recursion can terminate and produce a terminal word at any (or infinitely many) point(s) and that infinitely many of them are pairwise different. You could have a grammar with unbounded length of derivation, but only finitely many derivations terminate. And for non-regular grammars you could have infinitely many derivations of the same word.