Remark 1
Let I be a finite set of identifiers and let us consider
the alphabet
= I { + ,  ,*,/, ,(,)}. 
(1) 
Let L be the language consisting of arithmetic expressions
over .
We shall show that L cannot be recognized by FA.
First we observe that every word
w = w_{1}^{ ... }w_{} of L
is a well parenthesized expression that is
( i = 1^{ ... }  1)  w_{1}^{ ... }w_{i}  w_{1}^{ ... }w_{i} and  w =  w 
(2) 
where
 u denotes the number of occurrences
of the letter x in the word u.
The proof is easy by induction.
Let us assume that L can be recognized by a DFA
= (, S, s_{0}, F,)
with N states and initial state s_{0}.
Let us consider an arithmetic expression a starting with N opening
parentheses. There exists N states
s_{1}, s_{2},..., s_{N} such that
for every
i = 1^{ ... }N  1 we have
s_{i+1} = (s_{i}, x) with x = ( 
(3) 
There exists at least two values i_{1} and i_{2} of the index i
in the range
0^{ ... }N such that
s_{i1} = s_{i2}.
Hence by adding i_{2}  i_{1} opening parentheses in front of a
we can build another word b recognized by L.
However b is clearly not a well parenthesized expression
and thus it is not an arithmetic expression.
Figure 1:
Well parenthesized expressions cannot be recognized by FA.

Therefore the language of arithmetic expressions over
cannot be recognized by FA.