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I have the following Context free grammar where:

S → SS
S → hSm
S → ε

and the question I'm answering is:

"Prove by induction on n,that for all n ≥ 0, the string $h^n$$m^n$ has a derivation in this grammar of n + 1 steps."

Here's what I've tried proving:

Base step: When n = 0, we have $h^0$$m^0$ whereby it holds true since

it can be represented by the context free grammar.

Inductive step: Assume k is true for some integer k ≥ 0 hence we can

say that $h^k$$m^k$ holds. We now want to prove for k+1 with that

h^k+1 m^k+1. If we take k = 2, then can derive the string hhhmmm

which can also be represented by the context free grammar hence, we

have proven that the derivation of this grammar of n+1 steps is true.

I have doubts about my prove and I'm not certain if I'm proving it correctly. Would appreciate some feedback on what can be improved.

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For the base case, you should (at this point) be more explicit even for relatively trivial statements. Why is $h^0m^0$ representable by the grammar?

Your inductive case is just completely wrong structurally. You've assumed that $k$ is some arbitrary natural number, you can't then "take it to be $2$". The best you could do is say "if $k=2$ then ..." but then you'd still need to cover the $k\neq 2$ case. As it is, the antecedent and consequent of that sentence are almost entirely unrelated. How does "proving" the $k=2$ case imply anything about the $k=7$ case, say, let alone the arbitrary $k$ case? And here too, you should be more explicit how the strings are "representable by the grammar".

For the inductive case, you know, via the inductive hypothesis, that you have a derivation of $h^km^k$. You need to show how to build a derivation for $h^{k+1}m^{k+1}$ from it. Your inductive case should talk about how you are modifying/extending the derivation of $h^km^k$, and how those modifications/extensions are justified.

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