CS 390 Solutions to Test I

1. T, F, T, T, F.

2(a) tex2html_wrap_inline17
(b) tex2html_wrap_inline19

3. Basis Step: If tex2html_wrap_inline21 , then tex2html_wrap_inline23 .

Induction: Assume that tex2html_wrap_inline25 for an arbitrary string x.
Then for an arbitrary symbol b, tex2html_wrap_inline31 by the definition of the length. By the induction hypothesis, tex2html_wrap_inline25 , which is equal to tex2html_wrap_inline35 . Hence tex2html_wrap_inline37 .
On the other hand tex2html_wrap_inline39 by the application of the definition of the length. Hence tex2html_wrap_inline41 .

4. By noticing the two cycles at tex2html_wrap_inline43 , tex2html_wrap_inline45 is obtained.





S. Toida
Fri Jun 5 18:39:56 EDT 1998