Proving L(G) = L using induction and deriving a schema Consider the following grammar G:
S -> aS
  -> aTb
  -> a
T -> aTb
  -> a

How would I prove that L( G) = L where L = { a^m b^n | 0 <= n < m} in two steps as follows.
a) By induction that L( G) SubsetEq L considering the following equality:
0 <= |w|_b < |w|_a + |w|_S + |w|_T

  where w is any sentential form.

for this, i started with the basis step as follows: 
Basis: Let the lenght of the derivative = 1, therefore:

    S => aS,  |w|_b = 0, |w|_a = 1, |w|_S = 1, |w|_T = 0, and the inequality is true

    S => atb,   |w|_b = 1, |w|_a = 1, |w|_S = 0, |w|_T = 1, and the inequality is true

    S => a,   |w|_b = 0, |w|_a = 1, |w|_S = 0, |w|_T = 0, and the inequality is true

which proves that the basis step is true, but how would I go about proving the inductive step?
b) And that L SubsetEq L( G) by finding a derivation schema for 
      any w in L.
Ok, so with everyones help, I managed to get a bit of the solution for part b.
If x = a^m+n b^n and n > 0, then:
S => a^m-1 S using S -> aS
  => a^m Tb using S -> aTb
  => a^m+n-1 Tb^n using T -> aTb
  => a^m+nb^n using T -> a 
If x = a^n, then:
S = a^n-1 S using S -> aS
S = a^n using S -> a

But from here, I am not sure how to get the final solution in order to prove that L(G) = L. From what I understand, the final step should result in S=a^m b^n correct?

If $x=a^{m+n}b^n$ and $n>0$, then we can derive it by these steps from rules of G:


*

*Use rule $S\rightarrow aS$, $m-1$ times to derive $a^{m-1}S$.

*then use rule $S\rightarrow aTb$ to derive $a^{m}Tb$.

*then use rule $T\rightarrow aTb$, $n-1$ times to derive $a^{m+n-1}b^n$.

*and at last use rule $T\rightarrow a$ to derive $a^{m+n}b^n$


If $x=a^n$, then we can derive it by these steps from rules of G:


*

*Use rule $S\rightarrow aS$, $n-1$ times to derive $a^{n-1}$.

*then use rule $S\rightarrow a$ to derive $a^n$

