Here, we need to compute the reverse of a stringusing the given definition of R and B. R is the recursive definition in which the last symbol is made the first symbol and R is applied on the rest of the string. The base case here that of the empty symbol i.e \(\lambda\) which is given in part B.

1. In the first step, R is applied on "cubs" which makes the symbol "s" to be the first symbol and then recursively R will be applied on "cub"

2. Similar to step 1, R is applied on "cub" and after "s", the next symbol will be "b" i.e. "sb" . R will now be applied on "cu"

3. Again, after applying R on "cu" , we get the next symbol to be "u" i.e "sbu". R will now be applied on "c".

4. Since, "c" is the last symbol, an empty string symbol is inserted before "c" so that R can be applied. This makes it "\(\lambda\)c". Now R will be applied on this.

5. After applying R, we get "c" to be the next symbol i.e. "sbuc". Now we are left with \(\lambda\).

6. Apply part B on \(\lambda\) which will give \(\lambda\) again.

7. Since \(\lambda\) is the empty string, it can be removed. Therefore, remove it. The reversed string becomes. "sbuc".

\(cubs^R\)

\(= s(cub)^R\)

\(= sb(cu)^R\)

\(= sbu(c)^R\)

\(= sbu(\lambda c)^R\)

\(= sbuc(\lambda)^R\)

\(= sbuc(\lambda\)

\(= sbuc\)

1. In the first step, R is applied on "cubs" which makes the symbol "s" to be the first symbol and then recursively R will be applied on "cub"

2. Similar to step 1, R is applied on "cub" and after "s", the next symbol will be "b" i.e. "sb" . R will now be applied on "cu"

3. Again, after applying R on "cu" , we get the next symbol to be "u" i.e "sbu". R will now be applied on "c".

4. Since, "c" is the last symbol, an empty string symbol is inserted before "c" so that R can be applied. This makes it "\(\lambda\)c". Now R will be applied on this.

5. After applying R, we get "c" to be the next symbol i.e. "sbuc". Now we are left with \(\lambda\).

6. Apply part B on \(\lambda\) which will give \(\lambda\) again.

7. Since \(\lambda\) is the empty string, it can be removed. Therefore, remove it. The reversed string becomes. "sbuc".

\(cubs^R\)

\(= s(cub)^R\)

\(= sb(cu)^R\)

\(= sbu(c)^R\)

\(= sbu(\lambda c)^R\)

\(= sbuc(\lambda)^R\)

\(= sbuc(\lambda\)

\(= sbuc\)