Define a sequence of sets as follows:
$$
\begin{aligned}
& A_0=\{a, b\}, \\
& \text { and, for } k>0,
\end{aligned}
$$
$A_k=|(\alpha+\beta)|$ there are $i, j<k$ such that $\alpha \in A_i \& \beta \in A_j \mid$.
Thus,
$$
A_1=\{(a+a),(b+b),(a+b),(b+a)\},
$$
and
$$
\begin{gathered}
A_2 \text { contains }(a+(a+a)),(b+(a+b)),((b+a)+b), \\
((a+a)+(b+b)) \text {, etc. }
\end{gathered}
$$
These are just algebraic expressions; they have no values. Let $\left.A=\cup\left|A_k\right| k \in N a t\right\}$. By the above definition, if $\gamma \in A$, then either $\gamma$ is $a$ or $b$, or else there are $\alpha, \beta \in A$ such that $\gamma$ is $(\alpha+\beta)$. Define the rank of $\gamma$ as follows:
if $(\gamma$ is $a$ or $\gamma$ is $b)$, then $\operatorname{rank}(\gamma)=0$,
if $\gamma$ is $(\alpha+\beta)$, then $\operatorname{rank}(\gamma)=\operatorname{rank}(\alpha)+\operatorname{rank}(\beta)+1$.
For any $k \in N a t$, there is an element of $A$ with rank $k$. Use CVI on $\operatorname{rank}(\gamma)$ to prove that, in any element of $A$, the number of left parentheses equals the number of right parentheses. [Remark. Compare with the next problem.]