Prove the equivalence between a CFG and a Context free language I have to prove that the language $L=\{a^ib^j:2i=3j+1\}$ and the CFG G with the following rewrite rules:

$S\rightarrow a^2Tb$

$T\rightarrow a^3Tb^2 |\epsilon$

are equivalent to each other.

I'm going to prove this by saying $L(G)\subseteq L$ and $L \subseteq L(G)$. I was able to prove the first one. I should be able to prove the second one with induction on the length on a word $w$. But how can I do that? The induction basis should start at length 3, because the smallest word $w \in L$ is $aab$. But the second word has a length of 8. This means the stepsize in the induction proof is 5. I've never done an inductive proof with a stepsize different from 1. How can I do something like this as it's not possible to state in the induction step that "We assume that the theorem holds for $|w|=n$, now we are going to proof that it holds for $|w|=n+5$"?

UPDATE:
I think I found a solution to my own question:

Inductionbasis: Suppose $|w|=3$. Then $w$ must be $aab$ because that's the only word of length 3 that satisfies the requirements for the language L. But $aab$ is also an element of $L(G)$. The basis is OK.

Induction hypothesis: Assume that the theorem holds for all words $w`$ with length $|w`|=3+5n$. We are now going to proof that the theorem also holds for all words $w$ with length $|w|=3+5(n+1)$.

$w$  must be of the form $\alpha a^3b^2 \beta$ to be an element of $L$. Now consider the word $z=\alpha \beta$ with length $|z|=3+5n$. Then, by induction, we know that $z \in L \implies z \in L(G)$. Because $z \in L(G)$ it must hold that $z$ is of the form $a^nb^m$ with $2n=3m+1$. If we add the 3 $a$'s and 2 $b$'s from $w$ to this equation, we find that $2(n+3)=3(m+2)+1$. The following applies to this equation:

$2(n+3)=3(m+2)+1$

$\iff 2n+6=3m+6+1$

$\iff 2n+6=3m+1+6$

$\iff 2n+6=2n+6$

Thus, $w$ also satisfies $L(G)$. QED

Is this proof correct?
Since this is a very simple "linear" grammar, its language can be determined by looking at the possible derivations.

Clearly a successful derivation has to start with $S\to a^2Tb$, then repeats $k$ times the production $T\to a^3Tb^2$ and closes with $T\to \epsilon$. That will give strings of the form $a^{2+3k}b^{1+2k}$ with $k\ge 0$. (1)

Then we need to establish that these are exactly the strings of the form $a^ib^j$ with $2i=3j+1$. (2)

One direction is easy, strings of the form (1) satisfy (2) -- you solved that yourself -- by observing that 
$2(2+3k) = 3(1+2k)+1$.   

Your proof for the other direction sounds good, but it is not clear to me how one argues that the induction has step size $5$ (other than by trial and error).

The following would work. Start with (2) for natural numbers $i,j$.

This implies $3j+1$ is even (and non-negative).

Or, $3j$ is odd, so $j$ is odd, which means it is of the form $j=2k+1$ for $k\ge 0$.

Then $2i=6k+4$, or $i=3k+2$, and we are done.

This argument seems rather ad hoc. Note the similarity between this exercise and linear algebra. There are two different representations for a line in 2D: using vectors or equations. Here these are:


$\{ (2,1)+ k\cdot(3,2) \mid k\in \mathbb R \}$
$\{ (x,y) \mid 2x-3y=1 \}$


Unfortunately, I do not think linear algebra tricks directly work, as we here must restrict ourselves to integers and not the real numbers.