Is the empty string a terminal symbol?

I ask this question with regards to a grammar in Chosmky Normal Form. The definition states that the rules must be of the following forms:

• A $\rightarrow$ BC

• A $\rightarrow$ a

• S $\rightarrow$ $\epsilon$

And A,B and C are non-terminal symbols, B and C are different from S and a is a terminal symbol. The definition also states that the last rule can only be present if the language of the grammar accepts the empty string (source). Now if $\epsilon$ is a terminal symbol, than this is a contradiction as the second rule can be transformed to S $\rightarrow$ $\epsilon$, seeing as A must not be different from S. This leads me to conclude that $\epsilon$ is not a terminal symbol, is this correct?

It's true that, in general, definitions don't include the empty string in the set of "terminals", as there's no need for that (e.g. the production rules for a context-free grammar are defined as a relation $V \rightarrow (V \cup \Sigma)^*$ - the star covers all productions of the form $A \rightarrow \epsilon$; in all other contexts, it can be omitted because it has no effect on concatenation).

Note, however, that in the case of a CNF grammar, you couldn't just make a simple convention for "terminals" to include the empty string and cut down to just the first two rules, because rules of the form $A \rightarrow \epsilon$ are not allowed for all non-terminal $A$, just for $S$.

StackExchange Computer Science Q#85834, answer score: 7