so here a grammar R and a Langauge L, I want to prove that from R comes out L.
R={S→abS|ε} , L={(ab)n|n≥0}
so I thought I would prove that L(G) ⊆ L and L(G) ⊇ L are right.
for L (G) ⊆ L: I show by induction on the number i of derivative steps that after every derivative step u → w through which w results from u according to the rules of R, w = v1v2 or w = v1v2w with | v2 | = | v1 | and v1 ∈ {a} ∗ and v2 ∈ {b} ∗. and in the induction start: at i = 0 it produces that w is ε and at i = 1 w is {ε, abS}.
is that right so far ?
so here a grammar R and a Langauge L, I want to prove that from R comes out L.
Probably what you want to do is show that the language L(R) of some grammar R is the same as some other language L specified another way (in your case, set-builder notation with a regular expression).
so I thought I would prove that L(G) ⊆ L and L(G) ⊇ L are right.
Given the above assumption, you are correct in thinking this is the right way to proceed with the proof.
for L (G) ⊆ L: I show by induction on the number i of derivative steps that after every derivative step u → w through which w results from u according to the rules of R, w = v1v2 or w = v1v2w with | v2 | = | v1 | and v1 ∈ {a} ∗ and v2 ∈ {b} ∗. and in the induction start: at i = 0 it produces that w is ε and at i = 1 w is {ε, abS}.
This is hard for me to follow. That's not to say it's wrong. Let me write it down in my own words and perhaps you or others can judge whether we are saying the same thing.
We want to show that L(R) is a subset of L. That is, any string generated by the grammar R is contained in the language L. We can prove this by mathematical induction on the number of steps in the derivation of strings generated by the grammar. We start with the base case of one derivation step: S -> e produces the empty word, which is a string in the language L by choosing n = 0. Now that we have established the base case, we can state the induction hypothesis: assume that for all strings derived from the grammar in a number of steps up to and including k, those strings are also in L. Now we must prove the induction step: that any string derived in k+1 steps from the grammar is also in L. Let w be any string derived from the grammar in k+1 steps. From the grammar it is clear that the derivation of w must be S -> abS -> ababS -> ... -> abab...abS -> abab...abe = abab...ab. But this derivation is the same as the derivation of a string from the grammar in k steps, except that there was one extra application of S -> abS before the application of S -> e. By the induction hypothesis we know that the string w' derived in k steps is of the form (ab)^m for some m at least zero, and adding an extra application of S -> abS to the derivation adds ab. Because (ab)^m(ab) = (ab)^(m+1) we can choose n = m+1. So, all strings derived from the grammar in k+1 steps are also in the language, as required.
To prove that all strings in the language can be derived in the grammar, consider the following construction: to derive the string (ab)^n in the grammar, apply the production S -> abS a number of times equal to n, and the production S -> e exactly once. The first step gives an intermediate form (ab)^nS and the second step gives a closed form string (ab)^n.