NO i think only process change little bit
I have grammar
EX:- S->SA|A
A->a
Start from bottom A->a , gives FIRST(A)={a}
S->A , gives FIRST(S)=FIRST(A)={a}
S->SA , gives FIRST(S)=FIRST(S) , I think problem arises here. In such recursive calls rules says calculate FIRST(S) till it changes i.e. until elements are added in FIRST(S) continue to calculate. Once it stops changing that is you answer
Therefore FIRST(S)=FIRST(S)={a} , you call FIRST(S) as many times possible it won't change. Parsing Table:
So there are two entries for (S,a). Thus it is not LL(1