In G1, T can take the form of v * v * v * v * v ....
In G2, R can take the form * v * v * v ...,
so T can take the form of v * v * v * v ....
Thus, in both G1 and G2, E can take the form v * v * v + v * v * v * v + v * v ....
These are the algebraic expressions involving only addition and multiplication.
In G3, T can take the form v * v * v * v, so R can take the form + v * v + v * v. Thus, through the production E ! T R, E can take the form of any arithmetic expression.
However, E can also take the form of non-arithmetic expressions,
as with E --> R --> + T R --> + v R --> + v + T R ---> + v + v R --> + v + v.
In addition, G3 can generate ε , which G1 and G2 cannot.
Thus, while G1 and G2 generate the same language,
G3 generates a somewhat larger language. Incidentally, G1 demonstrates left-recursion because it contains productions of the form X --> X α ( in addition to non-problematic productions of the form X --> β ).
Left recursion can be eliminated by introducing a helper non-terminal R and changing the grammar to read
X --> β R
R --> α R | ε
hence C is the answer .
