Let $L$ be the set of all strings, including the null string, that can be constructed by repeated application of the following rules: If $\alpha\in L$ , then $a\alpha b\in L$ and $b\alpha a\in L$ . If $\alpha\in L$ and $\beta\in L$ , then $\alpha\beta\in L$ .

Prove that if $\alpha$ has equal number of $a$ 's and $b$ 's, then $\alpha\in L$ .

Solution

Verified

Step 1

1 of 3

Given:

$L$ =set of all strings including null string that can be constructed as follows:

$\bullet$ If $\alpha \in L$ , then $a\alpha b\in L$ and $b\alpha a\in L$ (1)

$\bullet$ If $\alpha \in L$ and $\beta \in L$ , then $\alpha \beta\in L$ (2)

To proof: If $\alpha$ has an equal number of $a$ 's and $b$ 's, then $\alpha\in L$ .

$\textbf{PROOF BY STRONG INDUCTION}$

Let $P(n)$ be the statement "If $\alpha$ has an equal number of $a$ 's and $b$ 's and of length $2n$ , then $\alpha\in L$ ".

$\textbf{Basis step}$ $n=0$

$\lambda$ is the only string of length 0.

$\lambda$ has an equal number of $a$ 's and $b$ 's (as it contains 0 $a$ 's and 0 $b$ 's).

However, $\lambda\in L$ by definition of $L$ and thus $P(0)$ is true.

$\textbf{Inductive step}$ Let $P(0),P(1),...,P(k)$ be true.

\text{If $\alpha$ has an equal number of $a$'s and $b$'s and of length $2i$, then $\alpha\in L$}

\text{ for }i=0,1,...,k

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