程序代写代做代考 When $n = 0$, $V_n = V_0 =STRING$, It has derivation $S \rightarrow E \rightarrow STRING$. The lenth of this derivation is 2. Thus, when $n = 0$, $V_n$ has a derivation of $n+2$.

When $n = 0$, $V_n = V_0 =STRING$, It has derivation $S \rightarrow E \rightarrow STRING$. The lenth of this derivation is 2. Thus, when $n = 0$, $V_n$ has a derivation of $n+2$.

Suppose for $k \ge 0$, $V_k$ has a derivation of $k+2$, we now show that $V_{k+1}$ has a derivation of $(k+1)+2$

$V_k$ has derivation $S \rightarrow E \Rightarrow V_{k} $

The total length is $k+2$, thus $E \Rightarrow V_{k}$ has length $k+1$.

$V_{k+1} = VIGENERE(V_{k}, STRING)$

We have derivation $S \rightarrow E \rightarrow VIGENERE(E , STRING) \Rightarrow VIGENERE(V_{k}, STRING)$

Because $E \Rightarrow V_{k}$ has length $k+1 $ , this derivation has length $(k+1)+2$.

From above, we can conclude that $V_n$ has a derivation of length $n+2$.