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For language $L = \{ x \in \{a,b\}^* \mid \#_a x = \#_b x \}$, I came up with the following CFG: $$S \rightarrow aSbS \mid bSaS \mid \varepsilon.$$ It can be easily shown that it is correct (quick example, for string $x = aabb$ we have these derivations: $S \rightarrow aSbS \rightarrow^* aaSbSb\varepsilon \rightarrow^* aa\varepsilon b \varepsilon b \rightarrow aabb$). Then I was asked to provide a mathematical proof, which is more like to be intended as a formalism, to prove, indeed, the correctness of such CFG.

Arguably, by induction, one could assume it holds that $\forall k\leq n$, $|x| = 2k$ and then test a string $x$ with number of characters $|x| = 2n + 2$. So string $x$ is in the form of $x = au'bu''$. But now how can I show that $u'$ and $u''$ still contain an equal number of $a$'s and $b$'s? Sorry if I'm missing something and might be easier than I think.

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There are two implications involved in the correctness. Every string in $L$ can be derived by the CFG (completeness) and every string derived by the grammar belongs to $L$ (consistency). I think you want an argument to formalize the first implication by induction.

Assuming the string $x\in L$ starts with an $a$, there must be a position with letter $b$ such that the string is of the form $aubv$ such that the $\#_a(u)=\#_b(u)$. The argument I know is by counting. Read the letters in $x$ one by one, and add $1$ for each $a$, subtract $1$ for each $b$. As the number of $a$'s equals the number of $b$'s, the count starts and stops with $0$. The counts becomes $1$ after the first letter, and must drop back to $0$ at some position with a $b$. In between we have seen an equal number of $a$'s and $b$'s.

Now that $x=aubv$ with $x\in L$ and $u\in L$ it is easy to argue that also $v\in L$.

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  • $\begingroup$ Good reasoning, and as regards the consistency property, i.e. $ L(G) \subseteq L $, it is basically already proved because, being $S \rightarrow aSbS \mid bSaS $, we can reach $ x = aubu$ or $ x = buau$ at a certain point, and $x \in L$, right? $\endgroup$ – Antonio Frighetto Oct 26 '18 at 11:37
  • $\begingroup$ @Antonio Indeed, $L(G)\subseteq L$ is easy by design of the grammar. If $S\Rightarrow^* u$ and $S\Rightarrow^* v$ assuming $u,v\in L$, then also $aubv, buav\in L$. Induction on the length of the derivation, or depth of the derivation tree. $\endgroup$ – Hendrik Jan Oct 26 '18 at 14:12

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