Indirect Left recursion

Question

I'm resolving indirect left recursion for these production rules:

S -> Aa / a   eq1
A -> Sb / b.  eq2

Where S is the starting symbol.

Now I can do this in two ways:

1. Putting A in eq1

   So I'll get the solution (sol1):

   S -> Sba /a /ba
   

   and then

   S -> aS' / baS' 
   S' -> baS' / epsilon
   

2. Replacing S in eq2:

   So I'll get the solution (sol2):

   S -> Aa / a 
   A -> abA' / bA'
   A' -> abA' / epsilon
   

Both seem to be correct. Which is the correct answer?

Answer 1

Yes, both your answers are correct. Note that the language under consideration is regular, see wikipedia.

The first grammar is equivalent to this regular expression: $$r_1 = a(ba)^* \mid (ba)^+.$$

The second grammar is equivalent to this regular expression: $$r_2 = (ab)^+a \mid b(ab)^*a \mid a.$$

It's easy to prove those regexes generate equal languages $L(r_1) = L(r_2)$.