What language does S -> A generates?

Question

I read formal grammar's definition from Wikipedia, and it seems like there can be such a grammar like:

S -> A

//no more rules

S and A are both non-terminals.

What language does it generates? I've thought several possibilities and rationales:

1. An empty set, because replacing S to A doesn't generates any actual strings which only contains terminals.
2. Any set can be an answer, because it's like solving equation S = A, to figure out what is S. Since there are no restrictions to set A, set A and set S can be anything.
3. This is not a valid grammar because it doesn't make any sense that A, which is non terminal, doesn't have any descriptions.

Which one is right?

Answer 1

To answer this, follow the definitions. From Wikipedia (see further references there):

• The language of a grammar is the set of sentences it builds.
• A sentence is a sentential form that contains no nonterminals.
• A sentential form is a member of (Σ ∪ N)*.
• Σ are terminal symbols, here ∅.
• N are nonterminal symbols, here {S, A}.

Working our way up this list again, sentential forms are members of {S, A}*, and sentences are members of ∅*. Since ∅* = {ε} (where ε is the empty string), the language may now be either ∅ or {ε}. But ε is not generated by the language, because there is no rule that generates it.

Once you are more comfortable with this kind of questions, you can also intuitively reason that since there is no production rule that generates only terminal symbols, the generated language must be ∅. However, that does not help you to see why the grammar is valid and why your third reasoning is not correct.

If you check the definitions carefully, you see that there is no restriction that every nonterminal symbol must have corresponding rewrite rules. (However, you can show that for each grammar with such nonterminal symbols there is a smaller grammar describing the same language.)